A) 13.834 kcal
B) 11.714 kcal
C) 12.726 kcal
D) 14.312 kcal
Correct Answer: C
Solution :
Given that, \[{{T}_{1}}=25{}^\circ C=298K\] \[{{T}_{2}}=35{}^\circ C=308K\] \[\frac{{{K}_{{{p}_{2}}}}}{{{K}_{{{p}_{1}}}}}=2R=2cal{{K}^{-1}}mo{{l}^{-1}}\] From vant Hoff equation \[2.303\log \frac{{{K}_{{{P}_{2}}}}}{{{K}_{{{p}_{1}}}}}=\frac{\Delta H}{R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] \[2.303\log 2=\frac{\Delta H}{2}\left[ \frac{308-298}{308\times 298} \right]\] or, \[\Delta H=12726.2\text{ }cal=10.726\text{ }kcal\]
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