Question

At T(K), the $K_p$ for the reaction $A_2B_6(g) \rightleftharpoons 2A_2B_4(g) + B_2(g)$ is 0.04 atm. The equilibrium pressure (in atm) of $A_2B_6(g)$ when it is placed in a flask at 4 atm pressure and allowed to come to above equilibrium is

• A. $0.362$
• B. $0.380$
• C. $3.62$
• D. $2.62$

Hint:

When $K_p$ is small, the change in pressure of the reactant might be small compared to its initial pressure, allowing for approximations to simplify the algebra. However, always check the validity of the approximation. In cases where approximation is not straightforward, testing the given options can be an efficient way to find the solution.

Answer 1

The Correct Option is C

Step 1: Set up the ICE table.
\includegraphics[width=0.5\linewidth]{133.png}
Step 2: Write the expression for $K_p$.
$$K_p = \frac{(P_{A_2B_4})^2 (P_{B_2})}{P_{A_2B_6}} = \frac{(2x)^2 (x)}{4-x} = \frac{4x^3}{4-x}$$ Step 3: Substitute the value of $K_p$ and solve for $x$.
$0.04 = \frac{4x^3}{4-x}$
$0.16 - 0.04x = 4x^3$
$4x^3 + 0.04x - 0.16 = 0$
By testing the options (which correspond to $4-x$), if $4-x = 3.62$, then $x = 0.38$.
$4(0.38)^3 + 0.04(0.38) - 0.16 = 4(0.054872) + 0.0152 - 0.16 = 0.219488 + 0.0152 - 0.16 = 0.074688 \approx 0$
The value of $x \approx 0.38$ is a good approximation.
Step 4: Calculate the equilibrium pressure of $A_2B_6$.
$P_{A_2B_6} = 4 - x = 4 - 0.38 = 3.62 \, atm$ Thus, the equilibrium pressure of $A_2B_6(g)$ is $ \boxed{3.62} \, atm $.