The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained. Concept Introduction: In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction. The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as, a A + b B ⇌ c C + d D The equilibrium constant for the above chemical reaction is expressed as, K = C c D d A a B b .

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Answer 1

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

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Answer 2

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

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Answer 3

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

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Answer 4

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

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Answer 5

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Answer 6

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Answer 7

Chapter 17, Problem 6ALQ

Interpretation Introduction

Interpretation:

The correct and incorrect things in the given statements are to be indicated. The incorrect statements are to be corrected and explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

The equilibrium constant of a reaction is expressed as the ratio of the concentration of the products and reactants, each raised to the power of their stoichiometric coefficients. The general equilibrium reaction is represented as,

$aA+bB⇌cC+dD$

The equilibrium constant for the above chemical reaction is expressed as,

$K=CcDdAaBb$ .

Answer 8

Expert Solution & Answer

Answer to Problem 6ALQ

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The correct statements are given below.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

The equilibrium constant for the given chemical reaction is calculated as follows:

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The given reaction is,

$A(g)+B(g)⇌C(g)$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B to the first statement, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

Therefore, the possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .

Hence, the second statement is incorrect.

The corrected statement is shown below.

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

Substitute the value of equilibrium concentration in the above equation

$K=CAB= 642= 0.75$

The new equilibrium constant for the given chemical reaction will be $0.75$ .

Hence, the first statement is correct and second statement is incorrect.

Answer 12

Conclusion

The equilibrium constant for the given chemical reaction is expressed as,

$K=CAB$

At equilibrium $A=2M, B=1M$ and $C=4M$ .

Put the value of equilibrium concentration in the above equation.

$K=CAB= 421= 2$

The equilibrium constant for the given chemical reaction is $2$ .

Hence, the first statement is correct.

It is given that after adding the $3 moles$ of B, the possible equilibrium condition is $A=1M, B=3M$ and $C=6M$ .

But after the addition of $3 moles$ of B, the concentration of B will increase to $4M$ and the concentration of the A will be same as $2M$ . Hence, the value of equilibrium constant will be change.

The possible equilibrium condition will be $A=2M, B=4M$ and $C=6M$ .