Question

For a gaseous phase reaction

When equilibrium is set up, $K =0.33$
Energy involved (in $kJ$ ) is

• A. 0.595
• B. 1.785
• C. 0.7854
• D. 1.5875

Answer 1

Answer: (A)

Cis $\rightleftharpoons$ Trans

Initial $1\,0$

At equilib. $(1-x) x$

$K=\frac{x}{1-x}=0.333=\frac{1}{3}$

$\therefore x=0.25$

Thus, there is $25 \%$ conversion into trans-form.

$\therefore \Delta_{r} H^{\circ}(\text { net })=2.38 \times 0.25$

$=0.595 \,KJ$