A) \[86600-\frac{\ln \left( 1.6\times {{10}^{12}} \right)}{R\left( 298 \right)}\]
B) \[0.5\left[ 2\times 86,600-R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right) \right]\]
C) \[R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right)-86600\]
D) \[86600+R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right)\]
Correct Answer: B
Solution :
\[\Delta {{G}_{R}}=-RT\ln Kp\] \[\Delta {{G}_{R}}=2{{\Delta }_{f}}{{G}_{N{{O}_{2}}}}-2\Delta {{G}^{0}}_{NO}\] \[-RT\ln 1.6\times {{10}^{12}}=2\left( {{\Delta }_{f}}{{G}_{N{{O}_{2}}}}-86,600 \right)\] \[86,600\frac{RT\ln \left( 1.6\times {{10}^{12}} \right)}{2}={{\Delta }_{f}}{{G}_{N{{O}_{2}}}}\] \[{{\Delta }_{f}}{{G}_{N{{O}_{2}}}}=0.5\left[ 2\times 86600-R\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right) \right]\]You need to login to perform this action.
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