Using data from Appendix C in the textbook, calculate ΔG∘ at 298 K.

asianadam

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10 years ago

Using data from Appendix C in the textbook, calculate ΔG∘ at 298 K.

Consider the reaction 3CH_4(g)→C_3H_8(g)+2H_2(g)

Question 1
Part A
Using data from Appendix C in the textbook, calculate ΔG∘ at 298 K.
Express your answer using four significant figures.

Question 2
Part B
Calculate ΔG at 298 K if the reaction mixture consists of 41 atm of CH_4, 0.012 atm of C_3H_8, and 2.3×10^−2 atm of H_2.

I don't understand the formula of ΔG∘

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padre

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Staff Member

Educator

10 years ago

Similar question was asked

Question

Consider the reaction: 3CH4(g) Rightwards Arrow

C3H8(g) + 2H2(g)
a. Calculate DGof at 298 K (From Appendix C DGo(CH4) = -50.8 KJ/mol, DGof(C3H8) = -23.47 KJ/mol)

Calculate DG at 298K if the reaction mixture consists of 40.0 atm of CH4, 0.0100 atm of C3H8(g), and 0.0180atm of H2.

Solution

formula for calculating Δ G

Δ G = -RTlnKp ............. (1)

Δ G = -23.47- (3*-50.8)v=128.93

calculate Kp
Kp = .01*.018^2/40^3

Kp = 5.06 *10^-11

put this value in equation (1)

Δ G = -RTlnKp

Δ G = -58.76 KJ

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Tralalalala Slight Smile

Big A.

wrote...

10 years ago

This is all wrong. When dealing with partial pressure you must use the formula:

ΔG = ΔGo + RT ln Q

You must find ΔGo, which the first part of the problem asks to solve.

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meria

wrote...

9 years ago

Part A:

ΔG(rxn) = ΔG(products) - ΔG(reactants)

ΔG (H2) = 0 kJ/mol
ΔG(C3H8) = -23.47 kJ/mol
ΔG(CH4) = -50.8 kJ/mol Leftwards Arrow

- multiply by 3 because reactant : 3CH4

ΔG(rxn) = (0 + (-23.47 kJ/mol)) - (3*(-50.8 kJ/mol))

ΔG(rxn) = 128.93 kJ

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bio_man

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meria

wrote...

9 years ago

Part B:

Use: ΔG = ΔG° + RTlnQ

R = 8.314 J/mol *k (constant)
T = 298 K
Q = (products)/(reactants)
= [(0.012 atm)(1.8*10^-2 atm)^2)] / [(41 atm)^3)]
Q = 5.64124 *10^-11
from part a you got ΔG° = 128.9 kJ

so ΔG = 128.9kJ + (8.314 J/mol *k)(298K)(ln5.64124 *10^-11)
(make sure after multiplying to convert from J to kJ in order to be able to add 128.9 kJ to get right ΔG)
ΔG = 70.46 kJ

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