Question

Part A: Kp = 2.7×1042
At 25 ∘C the reaction from Part A has a composition as shown in the table below.
Substance Pressure
(atm)
C2H2(g) 4.35
H2(g) 3.75
C2H6(g) 1.25×10−2
What is the free energy change, ΔG, in kilojoules for the reaction under these conditions?

Answer 1

Answer on Question #79400 – Chemistry – General Chemistry

Part A: $K_p = 2.7 \times 10^{-4}$

At $25^{\circ}C$ the reaction from Part A has a composition as shown in the table below.

Substance Pressure (atm):

$C_2H_2(g)$ 4.35

$H_2(g)$ 3.75

$C_2H_6(g) 1.25 \times 10^{-2}$

What is the free energy change, $\Delta G$, in kilojoules for the reaction under these conditions?

Solution:

$$2H_2 + C_2H_2 \rightarrow C_2H_6$$ $$K_p = \frac{[C_2H_6]_p}{[H_2]^2 [C_2H_2]} = \frac{1.25 \times 10^{-2}}{(3.75)^2 4.35} = 2.04 \times 10^{-4}$$ $$\Delta G = -RT(\ln K_p) = -(8.314\ \mathrm{J/mol\cdot K}) \times (298\ \mathrm{K}) \times \ln(2.04 \times 10^{-4})$$ $$\Delta G = 21048\ \mathrm{J/mol} = 21.05\ \mathrm{kJ/mol}$$