A) 1523.6 kJ \[{{\varepsilon }_{1}}-({{i}_{1}}+{{i}_{2}})R-{{i}_{1}}{{r}_{1}}=0\]
B) - 243.6 kJ\[{{\varepsilon }_{2}}-{{i}_{2}}{{r}_{2}}-{{\varepsilon }_{1}}-{{i}_{1}}{{r}_{1}}=0\]
C) - 120.0 kJ\[-{{\varepsilon }_{2}}-({{i}_{1}}+{{i}_{2}})R+{{i}_{2}}{{r}_{2}}=0\]
D) 553.0 kJ\[{{\varepsilon }_{1}}-({{i}_{1}}+{{i}_{2}})R+{{i}_{2}}{{r}_{2}}=0\]
Correct Answer: C
Solution :
Key Idea, Enthalpy of reaction \[\overrightarrow{E}=\hat{i}2xy+\hat{j}({{x}^{2}}+{{y}^{2}})+\hat{k}(3xz-{{y}^{2}})\] For reaction, \[\overrightarrow{E}=\hat{i}{{z}^{3}}+\hat{j}xyz+\hat{k}{{z}^{2}}\] \[\overrightarrow{E}=\hat{i}(2xy-{{z}^{3}})+\hat{j}x{{y}^{2}}+\hat{k}3{{z}^{2}}x\] \[2\times {{10}^{4}}J{{T}^{-1}}\] \[B=6\times {{10}^{4}}T\] \[60\Omega \] \[240\,\Omega \]You need to login to perform this action.
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