A) 1008 K
B) 1200 K
C) 845 K
D) 1118 K
Correct Answer: D
Solution :
\[\Delta S=\frac{\Delta H}{T}\] \[\Delta S=160.2\,J/K\] \[\Delta H=179.1\times {{10}^{3}}J/mol\] \[T=\frac{179.1\times {{10}^{3}}\,J/mol}{160.2\,\,J/M}\,=1117.9K\,\approx 1118K\]You need to login to perform this action.
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