A) 1
B) \[-1\]
C) 0
D) \[\pm \text{ }1\]
Correct Answer: B
Solution :
We have, \[\text{245 kJ mo}{{\text{l}}^{\text{-1}}}\] \[2HI(g){{H}_{2}}(g)+{{I}_{2}}(g)\] \[\text{1}.\text{4}\times \text{1}{{0}^{-\text{2}}}\] ...(i) \[\text{n}=\text{4}\] \[\text{n}=1\] where , \[\text{H=2}.\text{18}\times \text{1}{{0}^{-\text{18}}}\text{J at}{{\text{m}}^{\text{-1}}}\] \[h=6.625\times {{10}^{-34}}Js\] \[\text{1}.\text{54}\times \text{1}{{0}^{\text{15}}}{{\text{s}}^{-\text{1}}}\] \[\text{1}.0\text{3}\times \text{1}{{0}^{\text{15}}}\text{ J}{{\text{s}}^{-\text{1}}}\] \[3.08\times \text{1}{{0}^{\text{15}}}{{\text{s}}^{-\text{1}}}\] Clearly, \[\text{2}.0\times \text{1}{{0}^{\text{15}}}{{\text{s}}^{-\text{1}}}\] satisfies Eq. (i). \[H{{e}^{+}}\] \[{{n}_{2}}=3\xrightarrow{{}}{{n}_{1}}=1\]
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