Question

For the reaction \[\ce{A(g) <=> B(g)}\] at 495 K, ΔG° = −9.478 kJ mol⁻¹

If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ______ millimoles. (Round off to the Nearest Integer).

[R = 8.314 J mol⁻¹ K⁻¹; ln 10 = 2.303]

Options

• 10

• 15

• 20

• 25

Answer 1

For the reaction \[\ce{A(g) <=> B(g)}\] at 495 K, ΔG° = −9.478 kJ mol⁻¹

If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is 20 millimoles.

Explanation:

ΔG° = − RT ln K_eq

− 9.478 × 10³ = − 8.314 × 495 ln K_eq

ln K_eq = 2.303 = ln 10

So, K_eq = 10

Now,	A(g)	\[\ce{<=>}\]	B(g)
t = 0	22		0
t = t	22 − x		x

K_eq = `(["B"])/(["A"])`

10 = `"x"/((22 - "x"))`

x = 20

So, millimoles of B is 20.