Chemistry Reaction, Thermochemistry homework

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Description

Consider the reaction following reaction. Answer question B.

6 H2(g) + P4(g) → 4 PH3(g)
(a) Using data from Appendix C, calculate ΔG° at 298 K.
 
Given: 29.2 kJ

(b) Calculate ΔG° at 298 K if the reaction mixture consists of 7.8 atm of H20.049 atm of P4, and 0.24 atm of PH3.

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Explanation & Answer

Thank you for the opportunity to help you with your question!

for part 1;

rate of PH3 = 0.0027 / 1.5 = 0.0018 mol/L

rate of P4 = 1/4 * 0.0018 = 0.00045 mol/L

rate of H2 = 6/4 * 0.0018 = 0.0027 m

for part 2;

The dependence of free energy on pressure is given by the formula;

ΔG = ΔG° + RT ln Q

where Q = P^2(NH3) / P(N2) x P^3(H2)

Q = (0.65)^2 / (1.9)(1.6)^3
Q = 0.054

ΔG° = - 33.3 kJ/mol
T = 298 K
R = 8.3145 J/K.mol = 0.0083145 kJ/K.mol

ΔG = - 33.3 + (0.0083145)(298) ln(0.054)
ΔG = - 33.3 + (0.0083145)(298) (- 2.918)
ΔG = - 33.3 - 7.23 = - 40.53 kJ/mol

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Anonymous
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