A) \[1.8\times {{10}^{15}}\]
B) \[35\]
C) \[5.7\times {{10}^{-15}}\]
D) \[4.9\]
Correct Answer: C
Solution :
\[\Delta {{G}^{o}}=-RT\,\log \,{{K}_{p}}\] \[\Delta {{G}^{o}}=2.303\,RT\,\log \,{{K}_{p}}\] \[-\log \,{{K}_{p}}=\frac{\Delta {{G}^{o}}}{2.303RT}\] \[\because \] \[R=8.314,\] \[\Delta {{G}^{o}}=87\] \[-\log \,{{K}_{p}}=\frac{\Delta {{G}^{o}}}{2.303RT}\] \[T=273+25=298K\] \[-\log {{K}_{p}}=\frac{87}{2.303\times 8.314\times 298}\] \[\log {{K}_{p}}=-0.0153\] Antilog \[(-0.0153)=5.7\times {{10}^{-15}}\]
You need to login to perform this action.
You will be redirected in
3 sec
