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, \[K=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{(30)}^{2}}}{(10)(5)}\,=\frac{900}{50}=18\]
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