A) \[\overset{\bullet }{\mathop{OH}}\,\]
B) \[PC{{l}_{3}}(g)+C{{l}_{2}}(g)P{{C}_{5}}(g),\]
C) \[KI\]
D) \[{{C}_{6}}{{H}_{6}}+{{C}_{2}}{{H}_{5}}Cl\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}{{C}_{2}}{{H}_{5}}+HCl\]
Correct Answer: B
Solution :
We know that kinetic energy \[{{H}_{2}}\] where, k is a constant. Differentiating w.r.t t. we get \[{{I}_{2}}\] \[\log \frac{{{({{K}_{p}})}_{2}}}{{{({{K}_{p}})}_{1}}}=\frac{\Delta {{H}^{o}}}{2.303R}\left[ \frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}.{{T}_{2}}} \right]\] \[\log \frac{2.85}{1.40\times {{10}^{-2}}}=\frac{\Delta {{H}^{o}}}{2.303\times 8.3}\left[ \frac{400}{298\times 698} \right]\] But \[=2.3=\frac{400\Delta {{H}^{o}}}{3894250}\] \[\Delta {{H}^{o}}=22391.9=22.39kJ\]
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