Quiz 3 - Chemical Equilibrium | Chemical Equilibrium

The equilibrium constant K_eq​ for a reaction is the ratio of the rate constant for the forward reaction (k_f) to the rate constant for the reverse reaction (k_r​). This relationship holds because at equilibrium, the rates of the forward and reverse reactions are equal. Thus, the equilibrium constant is expressed as:

K_eq = $\frac{{\mathrm{k}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{r}}}$

This reaction has more gas molecules on the left side (3 moles) compared to the right side (2 moles). According to Le Chatelier's principle, if the volume of the system is decreased (which increases the pressure), the equilibrium will shift to the side with fewer moles of gas, which in this case is the side with NO₂. This shift will increase the partial pressure of NO₂ (g).

The equilibrium constant expression for this reaction is:

K_c = $\frac{{\left[{\mathrm{NO}}_2\right]}^2}{{\left[\mathrm{NO}\right]}^2\left[{\mathrm{O}}_2\right]}$

Now, substitute the given concentrations into the equation:

K_c = $\frac{{(0.18)}^2}{{(0.52)}^2 (0.24)}$ = $\frac{0.0324}{0.0649}$ = 0.50

Since the reaction produces more moles of gas on the right side (2 moles of NO₂) compared to the left side (1 mole of N₂O₄), increasing the volume decreases the pressure, causing the equilibrium to shift to the side with more gas molecules, thus increasing the amount of NO₂.

A decrease in volume increases pressure, favoring the side with fewer gas molecules. In option D, the reaction produces one mole of CO₂ (g), a gas, from a solid. Decreasing volume will shift the equilibrium to the left, reducing the amount of CO₂ (product) to minimize pressure.

Since the reaction is endothermic (ΔH^o > 0), lowering the temperature will shift the equilibrium toward the reactants (Ca(OH)₂) to produce more heat, thus increasing the amount of Ca(OH)₂ (s).

• Option A: Crushing does not affect the amount at equilibrium.
• Option C: Adding more CaO(s) does not shift the equilibrium because CaO is a solid and solids don't affect equilibrium concentrations.
• Option D: Adding N₂ (g) is irrelevant as it does not participate in the reaction.

In the reaction N₂ ​(g) + O_2​ (g) ⇌ 2 NO (g), the number of gas molecules on both sides is the same (2 moles of reactants and 2 moles of product). Increasing pressure doesn't change the equilibrium because there is no difference in the number of moles of gas.

In contrast, the other reactions have a different number of gas moles on each side, so increasing pressure would shift the equilibrium toward the side with fewer gas molecules, increasing the yield of products.

If the entire reaction is multiplied by a factor n, the equilibrium constant for the new reaction is the original equilibrium constant raised to the power of n:
 

For the second reaction, the original equation is doubled. Therefore, the equilibrium constant K_p​ is squared. The new K'_p​ is:

K'_p = (Kp)² = (8.2 x 10⁻⁴)² = 6.7 x 10⁻⁷

The expression for K_p​ is:

K_p = $\frac{{\mathrm{P}}_{{\mathrm{SO}}_3}}{{\mathrm{P}}_{{\mathrm{SO}}_2} {\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)}^{1/2}}$ 

Substitute the given partial pressures into the equation:

K_p = $\frac{0.332}{(0.562) {(0.101)}^{1/2}}$ = $\frac{0.332}{0.179}$ = 1.86

Increasing the temperature will shift the equilibrium toward the product side (since the reaction is endothermic), but increasing the volume will shift the equilibrium toward the the side with more moles of gas, which is the reactant side, thus decreasing the number of moles of PCl₅ (g). Therefore, only increasing the temperature will increase the number of moles of PCl₅ (g).