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Constant: R -8.314 J K mol-0.08314 dmbar Kmol-0.08206 dm atm K mol (3 Points) A container is divided into two equal compartm
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Answer #1

Solution:

Free energy of mixing is calculated as,

ΔGmix = n1 RT ln X1 + n2 RT ln X2

Given,

n1 = 2 mol H2

n2 = 4 mol N2

Mole fraction of H2 = X1 = 2 /2+4 = 2/6 = 1/3

Mole fraction of N2 = X2 = 4 /2+4 = 4/6 = 2/3

T = 25 + 273 = 298 K

R = 8.314 J K-1 mol-1

Therefore,

ΔGmix = 8.314 x 298 ( 2 x ln 1/3 + 4 x ln 2/3)

= 2477.572 x ( 2 x - 1.11 + 4 x - 0.40)

= 2477.572 ( - 2.22 - 1.60)

= - 9464.325 J = - 9.46 kJ

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