Chapter 12, Problem 71QAP

Consider the system

$\text{A}\left(g\right)+2\text{B}\left(g\right)+\text{C}\left(g\right)\rightleftharpoons \text{2D}\left(g\right)$at 25°C. At zero time, only A, B, and C are present. The reaction reaches equilibrium 10 min after the reaction is initiated. Partial pressures of A, B, and D are written as P_A, P_B, and P_D. Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required).

(a) P_D at 11 min P_D at 12 min.

(b) P_A at 5 min P_A at 7 min.

(c) K for the forward reaction K for the reverse reaction.

(d) At equilibrium, K Q.

(e) After the system is at equilibrium, more of gas B is added. After the system returns to equilibrium, K before the addition of B K after the addition of B.

(f) The same reaction is initiated, this time with a catalyst K for the system without a catalyst the system with a catalyst.

(g) K for the formation of one mole of D K for the formation of two moles of D.

(h) The temperature of the system is increased to 35°C. P_B at equilibrium at 25°C P_B at equilibrium at 35°C.

(i) Ten more grams of C are added to the system. P_B before the addition of C P_B after the addition of C.

Interpretation Introduction

(a)

Interpretation:

The partial pressure of D needs to be compared at time 11 min and 12 min.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 71QAP

Partial pressure of D gas at 11 min will be equal to (EQ) the partial pressure of D at 12 min.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

Thus, the partial pressure of each gas after 10 min will be equal. Therefore, partial pressure of D gas at 11 min will be equal the partial pressure of D at 12 min.

Interpretation Introduction

(b)

Interpretation:

The partial pressure of A needs to be compared at time 5 min and 7 min.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

Answer to Problem 71QAP

The partial pressure of A gas at 5 min will be greater than (GT) the partial pressure of A gas at 7 min.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

Since, reaction is moving in forward direction, the partial pressure of A decreases with increase in time.

Therefore, the partial pressure of A gas at 5 min will be greater than the partial pressure of A gas at 7 min.

Interpretation Introduction

(c)

Interpretation:

The value of K for the forward reaction needs to be compared with the reverse reaction.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The equilibrium constant for the direction of reaction is more in which it is moving. If the reaction is in forward direction, K for forward reaction is more and if it is moving in reverse direction, K for reverse reaction is more.

Answer to Problem 71QAP

The value of K for forward reaction is greater than (GT) K for backward or reverse reaction.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

Initially, the partial pressure of gas D is not given thus, the reaction is moving in forward direction.

Therefore, the value of K for forward reaction is greater than K for backward or reverse reaction.

Interpretation Introduction

(d)

Interpretation:

The relation between the value of K and Q at equilibrium needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

Answer to Problem 71QAP

The value of K is equal to (EQ) the value of Q or reaction quotient at equilibrium.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

Since, there is no change in the concentration of species takes place at equilibrium, the value of K is equal to the value of Q or reaction quotient at equilibrium.

Interpretation Introduction

(e)

Interpretation:

If B is added after equilibrium, the change in K needs to be determined.

Concept introduction:

According to Le Chatelier’s principle, if at equilibrium, any change in temperature, concentration or pressure is applied to a system, the shift in equilibrium takes place to counteract the change.

Also, on compressing a system, the total pressure of the system increases thus, reaction shifts to decrease the total pressure and number of moles of gaseous species.

Opposite to this on expansion, the total pressure of the system decreases thus, reaction shifts to increase the total pressure and number of moles of gaseous species.

Only change in temperature can change the value of K, in other cases the value of K remains the same.

There are 3 conditions that can change the equilibrium direction in a system:

1. Addition and removal of gaseous species.
2. Expansion and compression of the system.
3. Change in temperature of the system.

Answer to Problem 71QAP

The value of K before addition of B is less than (LT) K after the addition.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

After equilibrium, if more gas B is added, the reaction shifts in forward direction to decrease the partial pressure of gas B and the value of K increase.

Therefore, the value of K before addition of B is less than K after the addition.

Interpretation Introduction

(f)

Interpretation:

The value of K with or without catalyst needs to be compared.

Concept introduction:

Catalyst increases the rate of the reaction; it does not affect the amount of product formed in the reaction. Therefore, it does not change the value of equilibrium constant for a reaction.

Answer to Problem 71QAP

The value of K with or without the catalyst is equal. (EQ)

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

In the presence of catalyst, the value of K remains the same. The presence of catalyst only increases the rate of reaction, the amount of product formed remains the same thus, the value of K remains the same.

Therefore, the value of K with or without the catalyst is equal.

Interpretation Introduction

(g)

Interpretation:

If temperature of the system is increased after 20 min, the change in partial pressure of A needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Answer to Problem 71QAP

The value of K for formation of 1 mol of D will be less than (LT) the value of K for formation of 2 mol of D.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

The number of moles of gases is directly proportional to the partial pressure. If number of moles increases, partial pressure also increases.

For product, if partial pressure increases the value of K also increases thus, the value of K for formation of 1 mol of D will be less than the value of K for formation of 2 mol of D.

Interpretation Introduction

(h)

Interpretation:

The change in partial pressure of B due to increase in temperature at equilibrium needs to be determined.

Concept introduction:

According to Le Chatelier’s principle, if at equilibrium, any change in temperature, concentration or pressure is applied to a system, the shift in equilibrium takes place to counteract the change.

Also, on compressing a system, the total pressure of the system increases thus, reaction shifts to decrease the total pressure and number of moles of gaseous species.

Opposite to this on expansion, the total pressure of the system decreases thus, reaction shifts to increase the total pressure and number of moles of gaseous species.

Only change in temperature can change the value of K, in other cases the value of K remains the same.

There are 3 conditions that can change the equilibrium direction in a system:

1. Addition and removal of gaseous species.
2. Expansion and compression of the system.
3. Change in temperature of the system.

Answer to Problem 71QAP

The change in value of K or partial pressure of B cannot be determined due to increase in temperature.(MI)

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

If temperature of the system is increases, more information is required to determine the change in partial pressure of gas B.

The value of K depends on the temperature, but it also depends on the sign of the change in enthalpy of the reaction which depends on the type of reaction as if it is endothermic or exothermic reaction.

For exothermic reaction, the value of change in enthalpy is negative and for such reaction, the value of K decreases with increase in temperature.

And, for endothermic reaction, the value of change in enthalpy is positive and for such reaction, the value of K increases with increase in temperature.

Since, any information related to type of reaction or change in enthalpy, the change in value of K or partial pressure of B cannot be determined due to increase in temperature.

Interpretation Introduction

(i)

Interpretation:

The change in the partial pressure of A due to addition of 10 g of C to the system needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

For a general equilibrium reaction as follows:

$\text{A}\left(g\right)+\text{B}\left(g\right)\rightleftharpoons \text{C}\left(g\right)+\text{D}\left(g\right)$

The expression for the equilibrium constant is represented as follows:

$K=\frac{\left(P_C\right)\left(P_D\right)}{\left(P_A\right)\left(P_B\right)}$

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

Any species in pure solid and liquid phase does not participate in equilibrium expression thus, it does not affect the equilibrium of the reaction.

Answer to Problem 71QAP

The partial pressure of B after the addition of 10 g more solid C will be equal to (EQ) the partial pressure of B before the addition.

Explanation of Solution

The given reaction is as follows:

$A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)$

At zero time, only A, B and C is present.

After 10 min, equilibrium is reached.

If 10 g of more solid C is added to the system, there will be no change in the equilibrium takes place. This is because the species in solid phase cannot affect the value of K as they are not present in the equilibrium expression.

Thus, the partial pressure of B after the addition of 10 g more solid C will be equal to the partial pressure of B before the addition.