In the activity, click on the *K*eq and Δ*G*∘quantities to observe how they are related. Calculate Δ*G*∘ using this relationship and the equilibrium constant (*K*eq) obtained in Part A at *T*=298K:

*K*eq=1.92×1026

Express the Gibbs free energy (Δ*G*∘) in joules to three significant figures.

The following values may be useful when solving this tutorial.

Constant | Value |

E∘Cu | 0.337 V |

E∘Fe | -0.440 V |

R | 8.314 J⋅mol−1⋅K−1 |

F | 96,485 C/mol |

T | 298 K |

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- Josse
- Jul 20, 2016 at 12:40am

- But , G = - RT lnkeq Delt G = - 8.314 x 298 x ln 1.92^26 . Thus delt.G = -149941 J delta G = -149.9kJ
- Josse
- Jul 20, 2016 at 12:44am

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Delta G = - RT lnkeq delta G = - 8.314 x... View the full answer

Gibbs free energy (Δ G ∘) =... View the full answer