Question

At 700K, the value of $K_{c}$ for the following reaction is 64: $H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)$. If 0.5 mole of $H_{2}(g)$ and 0.5 mole of $I_{2}(g)$ were mixed at 700K and allowed to reach equilibrium, the amount of HI formed by mole is:

• A. 0.40
• B. 0.80
• C. 0.50
• D. 0.66

Hint:

When $\Delta n = 0$ (like in the $HI$ synthesis), $K_c = K_p = K_x$. In these cases, you don't need to know the volume of the container—you can work directly with moles!

Answer 1

The Correct Option is B

Concept:
The equilibrium constant \(K_c\) describes the ratio of product concentrations to reactant concentrations at equilibrium. We use an ICE (Initial, Change, Equilibrium) table to solve for equilibrium amounts. 

Step 1: Set up the ICE table
Let \(x\) be the moles of \(H_2\) reacted. 
\[ \begin{array}{c|c|c|c} & H_2 & I_2 & HI \\ \hline \text{Initial (mol)} & 0.5 & 0.5 & 0 \\ \text{Change (mol)} & -x & -x & +2x \\ \text{Equilibrium (mol)} & 0.5-x & 0.5-x & 2x \\ \end{array} \] 

Step 2: Apply equilibrium constant expression
Since \(\Delta n = 0\), volume cancels out. \[ K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(0.5-x)(0.5-x)} = 64 \] Taking square root: \[ \frac{2x}{0.5-x} = \sqrt{64} = 8 \] 

Step 3: Solve for \(x\)
\[ 2x = 8(0.5 - x) \] \[ 2x = 4 - 8x \] \[ 10x = 4 \Rightarrow x = 0.4 \] 
Final Answer:
\[ HI = 2x = 2 \times 0.4 = 0.80 \text{ moles} \]