At $\mathrm{T}(\mathrm{K})$, the equilibrium constant for $\mathrm{A}(\mathrm{g}) \rightleftharpoons…

At $\mathrm{T}(\mathrm{K})$, the equilibrium constant for $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})$ is $10^{-2}$. If rate of forward reaction is $0.025 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of backward reaction (in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ ) is
  1. $4 \times 10^1$
  2. $2.5 \times 10^{-4}$
  3. $2.5 \times 10^{-2}$
  4. None of the above

Solution

$\mathrm{K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}} \Rightarrow \mathrm{k}_{\mathrm{b}}=\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{c}}}=\frac{0.025}{10^{-2}}=2.5$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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