A) 0.52 V
B) 0.90 V
C) 0.30 V
D) 0.38 V
Correct Answer: A
Solution :
Key Idea Gibb's free energy is an additive property. \[\frac{d[N{{H}_{3}}]}{dt}=2\times {{10}^{-4}}\] For reaction, \[{{L}^{-1}}{{s}^{-1}},\] \[\frac{-d[{{H}_{2}}]}{dt}\] ...(i) For reaction, \[3\times {{10}^{4}}\,mol\,{{L}^{-1}}\,{{s}^{-1}}\] \[4\times {{10}^{4}}\,mol\,{{L}^{-1}}\,{{s}^{-1}}\] ...(ii) Adding Eqs. (i) and (ii), we get \[6\times {{10}^{4}}\,mol\,{{L}^{-1}}\,{{s}^{-1}}\]\[1\times {{10}^{4}}\,mol\,{{L}^{-1}}\,{{s}^{-1}}\] \[[O{{H}^{-}}]\] \[\text{Ba(OH}{{\text{)}}_{\text{2}}}\] \[TiF_{6}^{2-},CoF_{6}^{3-},C{{u}_{2}}C{{l}_{2}}\]\[NiCl_{4}^{2-}\]
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