Q31 / 200 · TS EAMCET 2002 · Chemistry

One mole of $A(g)$ is heated to $200^{\circ} \mathrm{C}$ in a one litre closed flask, till the following equilibrium is reached. $A(g) \quad B(g)$ The rate of forward reaction at equilibrium is $0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$. What is the rate (in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}$ ) of the backward reaction at equilibrium?

  • A
    $0.04$
  • B
    $0.01$
  • Correct answer
    C
    $0.02$
  • D
    $1$

Solution

At equilibrium, rate of forward reaction $=$ rate of backward reaction $\therefore$ rate of backward reaction at equilibrium $=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

Hence, the correct answer is option: 'C'.

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