Question

The reaction
$cis-Cr\left(en{)}_2\right(OH{)}_2^{+}\stackrel{k_1}{\rightleftharpoons }trans-Cr\left(en{)}_2\right(OH{)}_2^{+}$
is first order in both directions. At ${25}^{\circ }C$ the equilibrium constant is 0.16 and the rate constant $k_1$ is $3.3\times {10}^{-4}{s}^{-1}$. In an experiment starting with the pure cis form, how long would it take for half the equilibrium amount of the trans isomer to be formed ?

• A. 4.83 min
• B. 5.66 min
• C. 7.2 min.
• D. 9 min

Answer 1

The correct option is A 4.83 min

$k=0.16,k_1=3.3\times {10}^{-4}$
so $k_2=k_1/k=2.0625\times {10}^{-3}$
$k_1+k_2=0.0023925$
$[B{]}_{eq.}=\frac{k_1}{k_2}[A{]}_{eq.}=\frac{k_1[A{]}_0}{k_1+k_2}$
given $\left[B\right]=\frac{[B{]}_{eq.}}{2}=\frac{k_1[A{]}_0}{2(k_1+k_2)}$
& $\left[B\right]=\frac{k_1[A{]}_0}{k_1+k_2}[1-e^{-(k_1+k_2)t}]$
so $\frac{1}{2}=1-e^{-(k_1+k_2)t}$$(k_1+k_2)t=\ln 2$
$t=289.71sec.=4.82min.$