Question

A certain amount of H$_{2(g)}$ and I$_{2(g)}$ are sealed in a 4L container and kept at 600K to attain equilibrium. K$_C$ for the reaction, H$_{2(g)}$ + I$_{2(g)} \rightleftharpoons$ 2HI\(_{(g)}\) at 600K is 64. If the equilibrium concentration of HI\(_{(g)}\) is 0.08M, what are the equilibrium concentrations of H$_{2(g)}$ and I$_{2(g)}$ at the same temperature?

• A. 0.02 M
• B. 0.01 M
• C. 0.03 M
• D. 0.04 M

Hint:

To avoid big decimal math, take the square root of both sides of the equation first:
\(\sqrt{64} = \sqrt{\frac{[HI]^2}{x^2}} \implies 8 = \frac{0.08}{x} \implies x = \frac{0.08}{8} = 0.01\). This is much cleaner!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The equilibrium constant \(K_c\) is the ratio of the product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient.

Step 2: Key Formula or Approach:
For the reaction \(H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}\):
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

Step 3: Detailed Explanation:
1. Given: \(K_c = 64\) and \([HI] = 0.08 \text{ M}\).
2. Since H$_2$ and I$_2$ react in a 1:1 ratio and we started with "a certain amount" of each (assuming they were equal or effectively treating their concentrations as equal at equilibrium for the purpose of finding a shared value), let \([H_2] = [I_2] = x\).
3. Substitute the values into the \(K_c\) expression:
\[ 64 = \frac{(0.08)^2}{x \cdot x} \]
\[ 64 = \frac{0.0064}{x^2} \]
4. Solve for \(x^2\):
\[ x^2 = \frac{0.0064}{64} = 0.0001 \]
5. Solve for \(x\):
\[ x = \sqrt{0.0001} = 0.01 \text{ M} \]

Step 4: Final Answer:
The equilibrium concentrations of H$_{2(g)}$ and I$_{2(g)}$ are 0.01 M.