Question

Consider the system $$ \mathrm{A}(g)+2 \mathrm{~B}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{D}(g) $$ at \(25^{\circ} \mathrm{C}\). At zero time, only \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are present. The reaction reaches equilibrium 10 min after the reaction is initiated. Partial pressures of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{D}\) are written as \(P_{\mathrm{A}}, P_{\mathrm{B}},\) and \(P_{\mathrm{D}}\) 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) \(P_{\mathrm{D}}\) at \(11 \mathrm{~min}\) $$ P_{\mathrm{D}} \text { at } 12 \mathrm{~min} $$ (b) \(P_{\mathrm{A}}\) at \(5 \mathrm{~min} \longrightarrow P_{\mathrm{A}}\) at \(7 \mathrm{~min}\). (c) \(K\) for the forward reaction reaction. (d) At equilibrium, \(K \longrightarrow\) (e) After the system is at equilibrium, more of gas \(\mathrm{B}\) is added. After the system returns to equilibrium, \(K\) before the addition of \(\mathrm{B} \longrightarrow \mathrm{K}\) after the addition of \(\mathrm{B}\). (f) The same reaction is initiated, this time with a catalyst. \(K\) for the system without a catalyst \(K\) for the system with a catalyst. (g) \(K\) for the formation of one mole of \(\mathrm{D}\) \(K\) for the formation of two moles of \(\mathrm{D}\). (h) The temperature of the system is increased to \(35^{\circ} \mathrm{C} . P_{\mathrm{B}}\) at equilibrium at \(25^{\circ} \mathrm{C}\) \(P_{\mathrm{B}}\) at equilibrium at \(35^{\circ} \mathrm{C}\). (i) Ten more grams of \(\mathrm{C}\) are added to the system. \(P_{\mathrm{B}}\) before the addition of \(\mathrm{C} \longrightarrow P_{\mathrm{B}}\) after the addition of \(\mathrm{C}\).

Short answer

Answer: The equilibrium constant K is affected by temperature. Depending on the reaction's enthalpy, increasing the temperature might shift the equilibrium either towards the reactants or products. If the reaction is endothermic (positive enthalpy), increasing the temperature will shift the equilibrium toward the products, thus increasing K. If the reaction is exothermic (negative enthalpy), increasing the temperature will shift the equilibrium towards the reactants, thus decreasing K. In this case, we don't have enough information about reaction enthalpy to determine how K will be affected.

Step-by-step solution

Check the equilibrium time

Since the system reaches equilibrium at 10 min, we know the system is still at equilibrium at 11 min.

Compare \(P_D\) at 11 min with \(P_D\) at equilibrium

At 11 min, the system is still at equilibrium, therefore: $$ P_D \text{ at } 11 \text{ min} \text{ EQ } P_D \text{ at equilibrium} $$ #b) At 5 min and 7 min#

Check the equilibrium time

The system reaches equilibrium at 10 min, meaning it is not at equilibrium at 5 min or 7 min.

Compare \(P_A\) at 5 min and 7 min

Since the system is not at equilibrium at 5 or 7 min and the forward reaction is still occurring, the concentration of A will decrease over time: $$ P_A \text{ at } 5 \text{ min} \text{ GT } P_A \text{ at } 7 \text{ min} $$ #c) K for the forward reaction#

Identify the equilibrium expression

The equilibrium constant, K, for the forward reaction is given by the expression: $$ K = \frac{P_D^2}{P_A P_B^2} $$ The value of K can be obtained from experimental data or given information if available. In this case, we don't have enough information to determine the value of K. So the answer is: $$ \text{K for the forward reaction: MI} $$ #d) At equilibrium, K#

Identify the equilibrium condition

At equilibrium, the rate of the forward and reverse reactions are equal, and the concentrations of the reactants and products (partial pressures) do not change. Therefore, K remains constant at the given temperature (25 °C). The answer is: $$ \text{At equilibrium, K: EQ} $$ #e) Adding more B#

Identify the change

When more of B is added to the equilibrium system, the system is no longer in equilibrium, and it will shift to minimize the change according to Le Châtelier's principle.

Determine K before and after adding B

The equilibrium constant, K, is not affected by changes in the concentrations of the reactants or products. Hence: $$ K \text{ before } B \text{ is added: } EQ \text{ K after } B \text{ is added} $$ #f) With and without a catalyst#

Understand the catalyst effect

A catalyst only speeds up the reaction, but it does not change the equilibrium position or the equilibrium constant K. Thus: $$ K \text{ without a catalyst: } EQ \text{ K with a catalyst} $$ #g) Formation of one mole and two moles of D#

Identify the stoichiometry difference

The given equilibrium equation is: $$ \mathrm{A}(g)+2 \mathrm{~B}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{D}(g) $$ The formation of one mole of D will have half the stoichiometry coefficients. Let's denote the equilibrium constants as \(K_1\) and \(K_2\), respectively.

Determine the relationship between the equilibrium constants

The equilibrium constants for the formation of one mole of D (\(K_1\)) and two moles of D (\(K_2\)) are related as follows: $$ K_2 = \sqrt{K_1} $$ Therefore: $$ K_1 \text{ for one mole of } D \text{ GT } K_2 \text{ for two moles of } D $$ #h) Temperature change#

Identify the reaction type

We don't have enough information about reaction enthalpy to determine which way the equilibrium will shift. So we cannot compare \(P_B\) at equilibrium at 25 °C and 35 °C. The answer is: $$ P_B \text{ at 25 °C } \text{ MI } P_B \text{ at 35 °C} $$ #i) Adding more C#

Determine the effect of adding more solid C

If we add more C, which is a solid, to the system, the equilibrium does not change because it does not affect the partial pressures. Therefore: $$ P_B \text{ before } C \text{ is added: } EQ \text{ P_B after } C \text{ is added} $$