A) 0.389
B) 0.111
C) 0.788
D) 0,222
Correct Answer: C
Solution :
\[{{H}_{2}}+\] \[{{I}_{2}}\] \[\] \[2\,\,HI\] Initial pressure 5.0 5.0 0 At equilibrium (5.0 - p) (5.0 - p) 2p \[{{K}_{p}}=\frac{{{({{p}_{HI}})}^{2}}}{{{p}_{{{H}_{2}}}}\times {{p}_{{{I}_{2}}}}}=\frac{{{(2\,p)}^{2}}}{(0.5-p)\,(0.5-p)}\] or \[49=\frac{{{(2\,p)}^{2}}}{{{(0.5-p)}^{2}}}\] or \[\sqrt{49}=\frac{2\,p}{0.5-p}\] or \[7=\frac{2\,p}{0.5-p}\] or \[2p=7\,(0.5-p)\] or \[2p=3.5-7p\] or \[9p=3.5\] or \[9p=3.5\] or \[p=\frac{3.5}{9}=0.388\] Hence, \[{{p}_{HI}}=2p=2\times 0.388=0.788\]atmYou need to login to perform this action.
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