Question

At T(K), the following gaseous equilibrium is established. W + X \( \rightleftharpoons \) Y + Z The initial concentration of W is two times to the initial concentration of X. The system is heated to T(K) to establish the equilibrium. At equilibrium the concentration of Y is four times to the concentration of X. What is the value of \(K_c\)?

• A. 0.375
• B. 1.333
• C. 2.666
• D. 5.333

Hint:

1. Set up an ICE (Initial, Change, Equilibrium) table or list concentrations. 2. Use the given initial conditions to relate initial concentrations. 3. Use the given equilibrium condition to relate equilibrium concentrations or solve for the extent of reaction (\(\alpha\)). 4. Substitute equilibrium concentrations into the expression for \(K_c = \frac{[\text{Products}]}{[\text{Reactants}]}\) (raised to their stoichiometric coefficients).

Answer 1

The Correct Option is C

The equilibrium reaction is: W + X \( \rightleftharpoons \) Y + Z Let the initial concentration of X be \( C_0 \).
Then the initial concentration of W is \( 2C_0 \).
Initial concentrations of Y and Z are 0.
Let \( \alpha \) be the amount of X that reacts to reach equilibrium.
According to stoichiometry, \( \alpha \) amount of W also reacts, and \( \alpha \) amount of Y and \( \alpha \) amount of Z are formed.
Concentrations at equilibrium: [W]\(_{eq}\) = \( 2C_0 - \alpha \) [X]\(_{eq}\) = \( C_0 - \alpha \) [Y]\(_{eq}\) = \( \alpha \) [Z]\(_{eq}\) = \( \alpha \) Given at equilibrium, the concentration of Y is four times the concentration of X: [Y]\(_{eq}\) = 4 \(\times\) [X]\(_{eq}\) \( \alpha = 4 (C_0 - \alpha) \) \( \alpha = 4C_0 - 4\alpha \) \( 5\alpha = 4C_0 \) \( \alpha = \frac{4}{5}C_0 = 0.
8C_0 \).
Now, find the equilibrium concentrations in terms of \(C_0\): [W]\(_{eq}\) = \( 2C_0 - \alpha = 2C_0 - 0.
8C_0 = 1.
2C_0 \) [X]\(_{eq}\) = \( C_0 - \alpha = C_0 - 0.
8C_0 = 0.
2C_0 \) [Y]\(_{eq}\) = \( \alpha = 0.
8C_0 \) [Z]\(_{eq}\) = \( \alpha = 0.
8C_0 \) The equilibrium constant \( K_c \) is: \[ K_c = \frac{[\text{Y}]_{eq}[\text{Z}]_{eq}}{[\text{W}]_{eq}[\text{X}]_{eq}} \] \[ K_c = \frac{(0.
8C_0)(0.
8C_0)}{(1.
2C_0)(0.
2C_0)} \] \[ K_c = \frac{0.
8 \times 0.
8 \times C_0^2}{1.
2 \times 0.
2 \times C_0^2} \] The \( C_0^2 \) terms cancel out.
\[ K_c = \frac{0.
64}{0.
24} \] \[ K_c = \frac{64}{24} \] Divide by 8: \[ K_c = \frac{8}{3} \] As a decimal: \( 8 \div 3 = 2.
666.
.
.
\) So, \( K_c = 2.
666.
.
.
\) This matches option (3).