Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : mathrm A (mathrm g) rightleftharpoons mathrm B (mathrm g) + mathrm C (mathrm g)$$\mathrm {A} (\mathrm {g}) \rightleftharpoons \mathrm {B} (\mathrm {g}) + \mathrm {C} (\mathrm {g})$$ If p$p$ being the total pressure, K_p$K_p$ is the pressure equilibrium constant and alpha$\alpha$ is the degree of dissociation, then which of the following is true at equilibrium?

### Related Formula K_p = fracp_B cdot p_Cp_A$$K_p = \frac{p_B \cdot p_C}{p_A}$$ ### Core Logic Let us write down the dissociation dynamics for the reaction starting with a$a$ moles of mathrmA(g)$\mathrm{A(g)}$: beginarrayrcccc & mathrmA(g) & rightleftharpoons & mathrmB(g) & + & mathrmC(g) \\ textInitial (t=0): & a & & 0 & & 0 \\ textEquilibrium (t=eq): & a(1-alpha) & & aalpha & & aalpha endarray$$\begin{array}{rcccc} & \mathrm{A(g)} & \rightleftharpoons & \mathrm{B(g)} & + & \mathrm{C(g)} \\ \text{Initial (t=0):} & a & & 0 & & 0 \\ \text{Equilibrium (t=eq):} & a(1-\alpha) & & a\alpha & & a\alpha \end{array}$$ textTotal moles at equilibrium = a(1-alpha) + aalpha + aalpha = a(1+alpha)$$\text{Total moles at equilibrium} = a(1-\alpha) + a\alpha + a\alpha = a(1+\alpha)$$ ### Step 1: Calculate Partial Pressures The mole fractions (X_i$X_i$) are: - X_A = frac1-alpha1+alpha$X_A = \frac{1-\alpha}{1+\alpha}$ - X_B = fracalpha1+alpha$X_B = \frac{\alpha}{1+\alpha}$ - X_C = fracalpha1+alpha$X_C = \frac{\alpha}{1+\alpha}$ If the total pressure of the gas mixture at equilibrium is p$p$, the partial pressures are: - p_A = left( frac1-alpha1+alpha right) p$p_A = \left( \frac{1-\alpha}{1+\alpha} \right) p$ - p_B = left( fracalpha1+alpha right) p$p_B = \left( \frac{\alpha}{1+\alpha} \right) p$ - p_C = left( fracalpha1+alpha right) p$p_C = \left( \frac{\alpha}{1+\alpha} \right) p$ ### Step 2: Relate K_p to alpha and p Using the expression for K_p$K_p$: K_p = fracp_B cdot p_Cp_A = fracleft(fracalpha1+alpharight)p cdot left(fracalpha1+alpharight)pleft(frac1-alpha1+alpharight)p$K_p = \frac{p_B \cdot p_C}{p_A} = \frac{\left(\frac{\alpha}{1+\alpha}\right)p \cdot \left(\frac{\alpha}{1+\alpha}\right)p}{\left(\frac{1-\alpha}{1+\alpha}\right)p}$ K_p = fracalpha^2 p1-alpha^2 implies fracalpha^21-alpha^2 = fracK_pp$$K_p = \frac{\alpha^2 p}{1-\alpha^2} \implies \frac{\alpha^2}{1-\alpha^2} = \frac{K_p}{p}$$ Since K_p$K_p$ is strictly a function of temperature, it remains constant. Therefore, if total pressure p$p$ **increases**, the term fracK_pp$\frac{K_p}{p}$ decreases, which demands that the term fracalpha^21-alpha^2$\frac{\alpha^2}{1-\alpha^2}$ must decrease. This is only possible if the degree of dissociation **alpha$\alpha$ decreases**. ### Pattern Recognition Le Chatelier's Principle Shortcut: For reactions with Delta n_g > 0$\Delta n_g > 0$, raising the pressure pushes the system in the reverse direction to decrease the gas moles, which logically decreases the degree of dissociation alpha$\alpha$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium