Question

In a reversible reaction, the study of its mechanism says that both the forward and reverse reaction follows first order kinetics. If the half-life of forward reaction $\left(t_{1/2}\right)$f is $400$s and that of reverse reaction $\left(t_{1/2}\right)$ b is $250$s, the equilibrium constant of the reaction is:

• A. $1.3$
• B. $0.433$
• C. $0.625$
• D. $1.109$

Answer 1

Answer: (C)

$0.625$

$K=\frac{K_f}{K_b}$

$EquilibriumConstant=\frac{\frac{0.693}{t_{1/2\left(f\right)}}}{\frac{0.693}{t_{1/2\left(b\right)}}}$

$EquilibriumConstant=\frac{250}{400}$

$EquilibriumConstant=0.625$