Question 34 of 69intermediate🔧 ApplyNumerical3 marks

The reaction, CO(g)+3H2(g)⇌CH4(g)+H2O(g)CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2_2 and 0.02 mol of H2_2O and an unknown amount of CH4_4 in the flask. Determine the concentration of CH4_4 in the mixture. The equilibrium constant, KcK_c for the reaction at the given temperature is 3.90.

Correct Answer

The concentration of CH4CH_4 in the mixture is 0.05850.0585 mol/L.

Exercise: EXERCISES | Q: 6.34 | (Page 45)
For More Understanding

Explanation

To find the unknown concentration of CH4, we use the equilibrium constant expression for the given reaction. Since the volume of the flask is 1 L, the molar concentrations are equal to the given number of moles. By substituting the known equilibrium concentrations and the value of Kc into the expression, we can solve for the concentration of CH4.

Solution Steps

  1. Step 1: Write the balanced chemical equation and the equilibrium constant expression.

    CO(g)+3H2(g)⇌CH4(g)+H2O(g)CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g) Kc=[CH4][H2O][CO][H2]3K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}

  2. Step 2: Note the given equilibrium concentrations. Since the volume is 1 L, moles = concentration. [CO]=0.30[CO] = 0.30 M [H2]=0.10[H_2] = 0.10 M [H2O]=0.02[H_2O] = 0.02 M [CH4]=x[CH_4] = x M (unknown)

    Kc=3.90K_c = 3.90

  3. Step 3: Substitute the values into the equilibrium expression.

    3.90=x×0.020.30×(0.10)33.90 = \frac{x \times 0.02}{0.30 \times (0.10)^3}

  4. Step 4: Solve for xx.

    3.90=0.02x0.30×0.0013.90 = \frac{0.02x}{0.30 \times 0.001} 3.90=0.02x0.00033.90 = \frac{0.02x}{0.0003} x=3.90×0.00030.02x = \frac{3.90 \times 0.0003}{0.02}

    x=0.0585x = 0.0585 M

  5. Step 5: State the final concentration. The concentration of CH4CH_4 in the mixture is 0.05850.0585 mol/L.

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