Question

For the reaction A(g) + 2B(g) ↔ 2C(g) ΔGo = −34.4 kJ. What is the value of ΔG in kJ at 25oC when the pressure of A is 1.3 atm, pressure of B is 1.5 atm, and the pressure of C is 2.7? answers

Answer 1

Answer: The
`\Delta G` for the reaction is -32.130 kJ

Step-by-step explanation:

For the given chemical reaction:

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

The expression of
`K_p` for the given reaction:

`K_p=((p_(C))^2)/((p_(A))* (p_(B))^2)`

We are given:

`p_(A)=1.3atm\\p_(B)=1.5atm\\p_(C)=2.7atm`

Putting values in above equation, we get:

`K_p=((2.7)^2)/(1.3* (1.50)^2)\\\\K_p=2.5`

To calculate the Gibbs free energy of the reaction, we use the equation:

`\Delta G=\Delta G^o+RT\ln K_p`

where,

`\Delta G` = Gibbs free energy of the reaction = ?

`\Delta G^o` = Standard Gibbs' free energy change of the reaction = -34.4 kJ = -34400 J (Conversion factor: 1 kJ = 1000 J)

R = Gas constant =
`8.314J/K mol`

T = Temperature =
`25^oC=[25+273]K=298K`

`K_p` = equilibrium constant in terms of partial pressure = 2.5

Putting values in above equation, we get:

`\Delta G=-34400J+(8.314J/K.mol* 298K* \ln(2.5)\\\\\Delta G=-32129.82J=-32.130kJ`

Hence, the
`\Delta G` for the reaction is -32.130 kJ