1. Tardigrade
  2. Question
  3. Chemistry
  4. For a general reaction P(g) +O(g) leftharpoons +S(g), the specific rate constant is k(forward) = 2.0 × 10-3 mol-1L s-1 at a certain temperature. Reaction starts with equimolar amounts of P and Q. Reaching at equilibrium, it is observed that P is twice that of R. The specific rate constant for the backward reaction is

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Q. For a general reaction
$P_{(g)} +O_{(g)} \rightleftharpoons +S_{(g)}$, the specific rate constant is $k_{(forward)} = 2.0 × 10^{-3}\, mol^{-1}L\, s^{-1}$ at a certain temperature. Reaction starts with equimolar amounts of $P$ and $Q$. Reaching at equilibrium, it is observed that $P$ is twice that of $R$. The specific rate constant for the backward reaction is

Equilibrium

Solution:

image
$\Rightarrow x=\frac{a}{3}$
$K_{c}=\frac{x^{2}}{\left(a-x\right)\left(a-x\right)}, \frac{a^{2}}{9\times\left(2x\right)^{2}}=\frac{\left(3x\right)^{2}}{9\times4x^{2}}$
$K_{c}=\frac{1}{4}=0.25,$ but $K_{c}=\frac{k_{f}}{k_{b}}$
or $k_{b}=\frac{k_{f}}{K_{c}}=\frac{2\times10^{-3}}{0.25}=8\times10^{-3}$