Question

For the following endothermic reaction at equilibrium:

$\mathbf{2}{\mathbf{SO}}_{\mathbf{3}}\left(\mathbf{g}\right)\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

which of the following changes will increase the value of K?

a. increasing the temperature

b. decreasing the temperature

c. removing${\mathbf{SO}}_{\mathbf{3}}\left(\mathbf{g}\right)$(constant T)

d. decreasing the volume (constant T)

e. adding$\mathbf{Ne}\left(\mathbf{g}\right)$(constant T)

f. adding${\mathbf{SO}}_{\mathbf{2}}\left(\mathbf{g}\right)$(constant T)

g. adding a catalyst (constant T)

Short answer

1. Increase in temperaturewill increase the value of K.
2. Decrease in temperature will decrease the value of K.
3. When${\mathrm{SO}}_3$is removed, no effect in the value of K is observed.
4. Decrease in volume does not affect the value of K.
5. Addition of neon gas does not affect the value of K.
6. Addition of${\mathrm{SO}}_2$gas does not affect the value of K.
7. Addition of catalyst does not affect the value of K.

Step-by-step solution

Endothermic Reaction

For an endothermic reaction, the reaction is always favored towards the right. The reason is, that when a reaction occurs, the heat is utilized by the reactants to form products.

Explanation

The equation is as follows:

$2{\mathrm{SO}}_3\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

The dependence of rate constant on temperature is as follows:

$\frac{\ln K_{eq}}{dT}=\frac{\Delta H}{R{T}^2}$

a.

When the temperature is increased, the pressure increases, and the reaction will shift to the right. Thus, the increase in temperature will increase the value of K.

b.

When the temperature is decreased, the pressure decreases and the reaction will shift to the left. Thus, the decrease in temperature will decrease the value of K.

c.

When${\mathrm{SO}}_3$is removed, the state of equilibrium is disturbed so there is no question of an equilibrium constant. Thus, this condition does not affect the value of K.

d.

As long as the temperature is maintained, the decrease in volume will not affect the value of K. Thus, this condition does not affect the value of K.

e.

On addition of neon gases, the value of K remains unchanged since it is a noble gas. Thus, this condition does not affect the value of K.

f.

The only factor that affects the value of K is temperature and since the temperature is constant, the value of K remains unchanged. Thus, this condition does not affect the value of K.

g.

A catalyst will change the rate of a reaction but it will not shift the equilibrium. Thus, this condition does not affect the value of K.