The Δ H ∘ values for the given decomposition reaction has to be determined. Concept introduction: To calculate Δ H ∘ value we can use the following formula Δ H o reaction = ∑ Δ H o f ( products ) − ∑ Δ H o f ( reactants ) Where, Δ H ∘ = standard enthalpy change formation Δ H ∘ is inversely proportional to the decomposition of reaction. When Δ H ∘ increases decomposition of reaction decreases.

Question

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Answer 1

Question

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

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Answer 2

Question

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

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Answer 3

Question

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

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Answer 4

Question

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

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Answer 5

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

Answer 6

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Answer 7

Chapter 21, Problem 21.33QP

(a)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction has to be determined.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

Answer 8

(a)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $MgCO3(s)→MgO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoMgCO3=ΔHof(MgO)+ΔHof(CO2)−ΔHof(MgCO3)$

$ΔHoMgCO3=(1)(-601.8 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1112.9 kJ/mol) = 117.6 kJ/mol$

Answer 13

Conclusion

The $ΔH∘$ values for the given decomposition reaction was determined

$ΔH∘$ of $MgCO3$ 117.6 kJ/mol

Answer 14

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

Answer 15

(b)

Interpretation Introduction

Interpretation:

The $ΔH∘$ values for the given decomposition reaction and easy decomposed by heat has to be explained.

Concept introduction:

To calculate $ΔH∘$ value we can use the following formula

$ΔHoreaction=∑ΔHof(products)−∑ΔHof(reactants)$

Where, $ΔH∘$ = standard enthalpy change formation

$ΔH∘$ is inversely proportional to the decomposition of reaction.

When $ΔH∘$ increases decomposition of reaction decreases.

Answer 16

(b)

Expert Solution

Answer to Problem 21.33QP

The $ΔH∘$ values for the given decomposition is

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ easily decomposes than $CaCO3$ .

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

To determine $ΔH∘$ values for decomposition of $CaCO3(s)→CaO(s)+CO2(g)$

Substitute the suitable values in the following equation we get the answer as follows

$ΔHoCaCO3=ΔHof(CaO)+ΔHof(CO2)−ΔHof(CaCO3)$

$ΔHoCaCO3=(1)(-635.6 kJ/mol) + (1)(-393.5 kJ/mol) - (1)(-1206.9 kJ/mol) = 177.8 kJ/mol$

To explain which one easily decomposes.

$ΔH∘$ is inversely proportional to the decomposition of reaction.
When $ΔH∘$ increases decomposition of reaction decreases.
So, $MgCO3$ easily decomposes than $CaCO3$ .

Answer 21

Conclusion

The $ΔH∘$ values for the given decomposition reaction and which one gets easy decomposed was explained.

$ΔH∘$ of $CaCO3$ 177.8 kJ/mol

$MgCO3$ is easily decomposes than $CaCO3$

Answer 22

Ch. 21.3 - Prob. 1RCF Ch. 21.3 - Prob. 2RCF Ch. 21 - Prob. 21.1QP Ch. 21 - Prob. 21.2QP Ch. 21 - Prob. 21.3QP Ch. 21 - Prob. 21.4QP Ch. 21 - Prob. 21.5QP Ch. 21 - Prob. 21.6QP Ch. 21 - Describe with examples the chemical and...Ch. 21 - Prob. 21.8QP

Answer to Problem 21.33QP

Explanation of Solution

Answer to Problem 21.33QP

Explanation of Solution

Want to see more full solutions like this?

Chapter 21 Solutions