A) -11.488J
B) +11.488J
C) -11.488kJ
D) zero
Correct Answer: C
Solution :
Use the following equation, \[\Delta G=2303\,nRT\log \left( \frac{{{p}_{2}}}{{{p}_{1}}} \right)\] \[\therefore \] \[\Delta G=2.303\times 2\times (8.314\,J\,{{K}^{-1}}mo{{l}^{-1}})\] \[\times (300\,K)\times \log \left( \frac{0.1\,atm}{1\,atm} \right)\] = - 11488J = - 11.488 kJYou need to login to perform this action.
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