For the reaction,

the equilibrium constant is 0.16 and k₁ is 3.3 × 10^-4 s^-1. The experiment is started with pure cis form. The time taken for half the equilibrium amount of trans isomer to be formed is about

Concept:

Cis-Trans equilibrium reaction is first order reaction which can be mathematically expressed as \(ln \frac{a}{a-x}\) = kt

where, a= initial reactant, a-x= reactant present after time t

k= rate constant = k_f + k_b  = k₁ + k₂

Equilibrium constant, \(K_{eq}=\frac{k_{f}}{k_{b}}=\frac{k_{1}}{k_{2}}\)

where k_f and k_b are forward and backward rate constants

Explanation:

Given,

K_eq= 0.16, k₁ = 3.3 \(\times\) 10-4 s-1 

\(k_{2}=\frac{k_{1}}{K_{eq}}\) = 2.0625 \(\times\) 10^-3 s^-1

k= k₁+k₂ = 3.3 \(\times\) 10-4 + 2.0625 \(\times\) 10-3 

           = 23.92 x 10^-4

 \(ln \frac{a}{a-x}\) = kt -------------(1)

For cis-trans equilibrium reaction where half of trans form is formed, \(a-x=\frac{a}{2}\)

Thus putting values in equation 1, we get

ln 2 = 23.92 \(\times\) 10-4  t

Putting ln 2 = 0.693,  t = 290 s

Conclusion: -

The time taken for half the equilibrium amount of trans isomer to be formed is about 290 s. So the option 1 is correct.