A) 6.4
B) 0.02
C) 50
D) 4.6
Correct Answer: A
Solution :
Given, \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g);\] \[{{K}_{P}}=41\] \[{{K}_{p}}=\frac{p_{N{{H}_{3}}}^{2}}{{{p}_{{{N}_{2}}}}p_{{{H}_{2}}}^{3}}=41\] ?(i) For reaction, \[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\rightleftharpoons N{{H}_{3}}(g),\] \[K_{p}^{'}=\frac{{{p}_{N{{H}_{3}}}}}{p_{{{N}_{2}}}^{1/2}p_{{{H}_{2}}}^{3/2}}\] ?(ii) On squaring both sides, we get \[{{(K{{'}_{p}})}^{2}}=\frac{p_{N{{H}_{3}}}^{2}}{{{p}_{{{N}_{2}}}}p_{{{H}_{2}}}^{3}}\] ?(iii) On dividing Eq. (i) by Eq. (iii), we get \[\frac{{{K}_{p}}}{{{(K'{{ & }_{p}})}^{2}}}=1\] or \[{{K}_{p}}={{(K{{'}_{p}})}^{2}}\] or \[K{{'}_{p}}=\sqrt{{{K}_{p}}}=\sqrt{41}=6.4\]
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