Question

\( \text{X}_2 + \text{O}_2 \rightleftharpoons 2\text{XO} \)
Concenttration of \( \text{X}_2 \) and \( \text{O}_2 \) are \( 4\times 10^{-3} \) and \( 3\times 10^{-3} \) respectively. Equilibrium concentration of XO (Kc = 0.5)

Hint:

When dealing with equilibrium constant expressions, pay strict attention to the stoichiometric coefficients. The coefficient '2' for XO means its concentration must be squared in the numerator.

Answer 1

Step 1: Understanding the Concept:
The equilibrium constant \( K_c \) for a reversible reaction is defined as the ratio of the product of the equilibrium concentrations of the products to the product of the equilibrium concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Step 2: Key Formula or Approach:
For the reaction \( \text{X}_2 + \text{O}_2 \rightleftharpoons 2\text{XO} \), the equilibrium expression is:
\[ K_c = \frac{[\text{XO}]^2}{[\text{X}_2][\text{O}_2]} \] Step 3: Detailed Explanation:
We are given:
\( K_c = 0.5 \)
\( [\text{X}_2] = 4 \times 10^{-3} \text{ M} \)
\( [\text{O}_2] = 3 \times 10^{-3} \text{ M} \)
We need to find the concentration of XO, let's denote it as \( [\text{XO}] \).
Substitute the given values into the \( K_c \) expression:
\[ 0.5 = \frac{[\text{XO}]^2}{(4 \times 10^{-3}) \times (3 \times 10^{-3})} \] Calculate the denominator:
\[ (4 \times 10^{-3}) \times (3 \times 10^{-3}) = 12 \times 10^{-6} \] Now the equation becomes:
\[ 0.5 = \frac{[\text{XO}]^2}{12 \times 10^{-6}} \] Multiply both sides by \( 12 \times 10^{-6} \):
\[ [\text{XO}]^2 = 0.5 \times 12 \times 10^{-6} \] \[ [\text{XO}]^2 = 6 \times 10^{-6} \] Take the square root of both sides to find \( [\text{XO}] \):
\[ [\text{XO}] = \sqrt{6 \times 10^{-6}} \] \[ [\text{XO}] = \sqrt{6} \times 10^{-3} \] We know that \( \sqrt{4} = 2 \) and \( \sqrt{9} = 3 \), so \( \sqrt{6} \) is approximately 2.45.
\[ [\text{XO}] \approx 2.45 \times 10^{-3} \text{ M} \] Step 4: Final Answer:
The equilibrium concentration of XO is \( 2.45 \times 10^{-3} \text{ M} \).