Chapter 12, Problem 69QAP

Consider the reaction:

$\text{A}\left(g\right)+2\text{B}\left(g\right)+\text{C}\left(s\right)\rightleftharpoons 2\text{D}\left(g\right)$At 25°C, only A, B, and C are present. The partial pressures of A, B, and D are given as P_A, P_B, and P_D. Equilibrium is established 18 minutes after the reaction starts. Use terms is less than (LT), is greater than (GT), is equal to (EQ), or insufficient information (X) to answer the following questions.

(a) P_D at 5 min P_A.

(b) P_A at 21 min P_A at 27 min.

(c) P_B at 7 min P_B at 13 min.(d) After 20 min, more B is added. When equilibrium is reestablished, K before the addition K after the addition.

(e) After 20 min, the temperature is increased to 35°C. P_A before the temperature increase P_A after the temperature increase after equilibrium is reestablished.

Interpretation Introduction

(a)

Interpretation:

The relation between the partial pressure of D and A at 5 min 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 69QAP

Partial pressure of D gas at 5 min is greater than (GT) partial pressure of A gas.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at $t+18$ min.

The change in partial pressure takes place as follows:

$\begin{array}{l}A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)\\ P_A{P}_B\text{ - -}\\ P_A\text{-}P{P}_B\text{-2}P\text{ - 2}P\end{array}$

At 5 min, the equilibrium is not established thus, the reaction is moving in the forward direction with same rate.

Since, the reaction is in forward direction, the pressure of product will be more than reactant.

Thus,

Partial pressure of D gas at 5 min is greater than partial pressure of A gas.

Interpretation Introduction

(b)

Interpretation:

The relation between the partial pressure of A 12 min and at 27 min 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 69QAP

The partial pressure of A gas at 21 min will be equal to (EQ) the partial pressure of B gas at 27 min.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at $t+18$ min.

The change in partial pressure takes place as follows:

$\begin{array}{l}A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)\\ P_A{P}_B\text{ - -}\\ P_A\text{-}P{P}_B\text{-2}P\text{ - 2}P\end{array}$

Since, equilibrium is established at 18 min, after 18 min the pressure of all the gases become equal.

Thus, the partial pressure of A gas at 21 min will be equal to the partial pressure of B gas at 27 min.

Interpretation Introduction

(c)

Interpretation:

The relation between the partial pressure of B at 7 min and at 13 min 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 69QAP

The partial pressure of A gas at 7 min will be greater than (GT) partial pressure of B gas at 13 min.

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at $t+18$ min.

The change in partial pressure takes place as follows:

$\begin{array}{l}A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)\\ P_A{P}_B\text{ - -}\\ P_A\text{-}P{P}_B\text{-2}P\text{ - 2}P\end{array}$

The reaction is moving in forward direction, the partial pressure of B decreases with time till the system reaches equilibrium. Thus, partial pressure of A gas at 7 min will be greater than partial pressure of B gas at 13 min.

Interpretation Introduction

(d)

Interpretation:

If more B is added in the system after 20 min, 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 69QAP

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

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at $t+18$ min.

The change in partial pressure takes place as follows:

$\begin{array}{l}A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)\\ P_A{P}_B\text{ - -}\\ P_A\text{-}P{P}_B\text{-2}P\text{ - 2}P\end{array}$

After 20 min, the system is in equilibrium and if more B is added the equilibrium will shift in the direction to decrease the partial pressure of B and the reaction moves in forward direction.

Thus, after the addition of B, K increases and the value of K before the addition of B will be less than K after the addition.

Interpretation Introduction

(e)

Interpretation:

If temperature of the system is increased after 20 min, the change in partial pressure of A 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 69QAP

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

Explanation of Solution

The reaction is given as follows:

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

Initially A, B and C are present in the system. If reaction starts at time t, the equilibrium will be established at $t+18$ min.

The change in partial pressure takes place as follows:

$\begin{array}{l}A\left(g\right)+2B\left(g\right)+C\left(s\right)\rightleftharpoons 2D\left(g\right)\\ P_A{P}_B\text{ - -}\\ P_A\text{-}P{P}_B\text{-2}P\text{ - 2}P\end{array}$

After 20 min, if temperature is increased to $35{}^{\circ }\text{C}$, the change in pressure can be determined due to insufficient information.

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 A cannot be determined due to increase in temperature.