question_answer

For the reaction: \[{{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI,\] if initially 25 mL of \[{{H}_{2}}\] and 20 mL of \[{{I}_{2}}\] are present in a container and at equilibrium, 30 mL of \[HI\] is formed, then calculate equilibrium constant.

Answer:

                                \[{{H}_{2}}\,\]     +                 \[{{I}_{2}}\,\]                 \[\rightleftharpoons\]                 \[2\,HI\] Initial:                 \[25\,mL\]                 \[20\,mL\]                                       \[0\] At eqm. :                 \[25-x\]                 \[20-x\]                               \[2\,x\]                                                 But \[2\,x=30\,mL\] (Given), i.e., \[x=15\] \[\therefore \] At equilibrium, \[{{H}_{2}}=25-15=10\,mL,\]\[{{I}_{2}}=20-15=5\,mL\] and \[HI=30\,mL\]. As equal volumes contain equal number of moles, therefore, volumes can be used in place of moles. Hence, \[K=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{(30)}^{2}}}{(10)(5)}\,=\frac{900}{50}=18\]