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tensa zangetsu [6.8K]
3 years ago
12

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 8.00 atm, PB = 5.40 atm, PC = 6.90

atm, and PD = 5.90 atm.
A(g) + 2B(g) -----> C(g) + D(g)

What is the standard change in Gibbs free energy of this reaction at 25 degrees Celsius
Chemistry
1 answer:
Yuliya22 [10]3 years ago
3 0

<u>Answer:</u> The standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

A(g)+2B(g)\rightleftharpoons C(g)+D(g)

The expression of K_p for the given reaction:

K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}

We are given:

p_A=8.00atm\\p_B=5.40atm\\p_C=6.90atm\\p_D=5.90atm

Putting values in above equation, we get:

K_p=\frac{6.90\times 5.90}{8.00\times (5.40)^2}\\\\K_p=0.174

To calculate the standard Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_p

where,

\Delta G^o = standard Gibbs free energy

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 0.174

Putting values in above equation, we get:

\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.174)\\\\\Delta G^o=4332.5J/mol=4.33kJ/mol

Hence, the standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

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