A) \[86600-\frac{in\left( 1.6\times 1{{0}^{12}} \right)}{R\left( 298 \right)}\]
B) \[0.5\left[ 2\times 86,600-R\left( 298 \right)In\left( 1.6\times 1{{0}^{12}} \right) \right]\]
C) \[\operatorname{R}\left( 298 \right)in\left( 1.6\times 1{{0}^{12}} \right)-86600\]
D) \[86600+R\left( 298 \right)in\left( 1.6\times 1{{0}^{12}} \right)\]
Correct Answer: B
Solution :
| \[\Delta \operatorname{G}_{NO\left( g \right)}^{{}^\circ }=86.6\operatorname{KJ}/\operatorname{mol}=86600\operatorname{J}/mol\] |
| \[\operatorname{G}_{N{{O}_{2}}(g)}^{{}^\circ }=xJ/mol\] |
| \[\operatorname{T}=298,{{K}_{P}}=1.6\times 1{{0}^{12}}\] |
| \[\Delta \operatorname{G}{}^\circ =-RT\,ln\,{{K}_{\operatorname{P}}}\] |
| Give equation, \[2NO\left( g \right)+{{O}_{2}}\left( g \right)2{{\operatorname{NO}}_{2}}\left( g \right)\]\[\therefore \,2\Delta \operatorname{G}_{N{{O}_{2}}}^{{}^\circ }-2\Delta \operatorname{G}_{NO}^{{}^\circ }=-\operatorname{R}\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right)\]\[2\Delta G_{N{{O}_{2}}}^{{}^\circ }-2\times 86600\] |
| \[=-R\left( 298 \right)\ln \,\left( 1.6\times {{10}^{12}} \right)\] |
| \[2\Delta G_{N{{O}_{2}}}^{{}^\circ }=2\times 86600-\operatorname{R}\left( 298 \right)\ln \left( 1.6\times {{10}^{12}} \right)\] \[\begin{align} & \Delta G_{N{{O}_{2}}}^{{}^\circ }=\frac{1}{2}[2\times 86600-R\left( 298 \right) \\ & \ln \,\left( 1.6\times {{10}^{12}} \right)] \\ \end{align}\] |
| \[=0.5[2\times 86600-R\left( 298 \right)\ln \,\left( 1.6\times {{10}^{12}} \right)\] |
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