A) \[-\text{14}.\text{6 kJ mo}{{\text{l}}^{\text{-1}}}\]
B) \[-\text{16}.\text{8 kJ mo}{{\text{l}}^{\text{-1}}}\]
C) \[+\text{16}.\text{8 kJ mo}{{\text{l}}^{-}}^{\text{1}}\]
D) \[+\text{244}.\text{8 kJ mo}{{\text{l}}^{\text{-1}}}\]
Correct Answer: C
Solution :
\[\frac{1}{2}{{l}_{2}}(S)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow[{}]{{}}lCl(g)\] \[\Delta H=\left[ \frac{1}{2}{{H}_{s\to g}}+\frac{1}{2}\Delta {{\Eta }_{diss}}(C{{l}_{2}})+\frac{1}{2}\Delta {{H}_{diss}}({{l}_{2}}) \right]\] \[\left( =\frac{1}{2}\times 62.76+\frac{1}{2}\times 242.3+\frac{1}{2}\times 151.0 \right)-211.3\] \[=228.03-211.3\] \[\Delta H=16.73\]You need to login to perform this action.
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