## Joe’s Review Session Last Problem

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

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Kellina Tran 2I
Posts: 50
Joined: Sat Jul 22, 2017 3:00 am

### Joe’s Review Session Last Problem

Can someone who went to Joe’s review session tonight give me a breakdown of the steps to calculate the last problem calculating deltaG using numbers as well please? Thank you!

Gwyneth Huynh 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Re: Joe’s Review Session Last Problem

He asked us to find the deltaG for the formation of NH3 (assuming standard conditions) and gave us:
deltaHf (NH3)= -46.11 kJ/mol
deltaSm (NH3)= 192.45 J/K*mol
deltaSm (H2)= 130.68 J/K*mol
deltaSm (N2) = 191.61 J/K*mol

We would use the reaction:
N2 (g) + 3H2 (g) --> 2 NH3 (g)

Since we're finding deltaG, we would use the equation, deltaG = deltaH - T(deltaS)
To find delta H, I took the enthalpy of formation of NH3 and multiplied that by 2, since the reaction results in 2 moles of NH3. Therefore, deltaH of the reaction is -92.22 kJ.

To find deltaS, I used the equation, [sum of n(standard molar entropies of the products)] - [sum of n(standard molar entropies of the reactants)].

Since the the reaction occurs under standard conditions, T = 298 K

Therefore, by plugging your values into the equation, deltaG = deltaH - TdeltaS, you should get around -33.99 kJ, which you will then divide by the number of moles of NH3 produced in the reaction (2 moles) to get -16.5 kJ/ mol.

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