MCQ
In a chemical equilibrium, the rate constant of the backward reaction is $7.5 \times {10^{ - 4}}$ and the equilibrium constant is $1.5$. So the rate constant of the forward reaction is
  • A
    $5 \times {10^{ - 4}}$
  • B
    $2 \times {10^{ - 3}}$
  • $1.125 \times {10^{ - 3}}$
  • D
    $9.0 \times {10^{ - 4}}$

Answer

Correct option: C.
$1.125 \times {10^{ - 3}}$
(c) ${K_c} = \frac{{{K_f}}}{{{K_b}}}$

${K_f} = {K_c} \times {K_b} = 1.5 \times 7.5 \times {10^{ - 4}}$$ = 1.125 \times {10^{ - 3}}$

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