A) \[-3.12\times {{10}^{4}}J\]
B) \[-1.25\times {{10}^{5}}J\]
C) \[25.0\times {{10}^{6}}J\]
D) None of these
Correct Answer: A
Solution :
[a] \[{{W}_{max}}=-n.FE;\] \[{{W}_{max}}=-2\times 96500\times 0.65=-1.25\times {{10}^{5}}J\] \[0.5g\,{{H}_{2}}=0.25mole\] Hence \[{{W}_{max}}\] \[=-1.25\times {{10}^{5}}\times 0.25=-3.12\times {{10}^{4}}J\]You need to login to perform this action.
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