Question

Hydrogen gas (H₂), bromine gas (Br₂) and hydrogen bromide gas (HBr) are added to a 2.0 L flask such that the concentration of each are initially 1.5 *10^-2 M H₂, 1.5*10^-2 M Br₂, and 5.0 *10^-1 M HBr.

H_2(g) + Br_2(g) ⇌ 2HBr_(g) K_c=6.25*10¹ at 35 degrees C.

a.) Calculate the concentration of HBr once the system reaches equilibrium.

b.) calculate delta G standard for the above reaction.

The K_c value is correct as listed

Answer 1

a)

[H2] = 0.0535M

[Br2] = 0.0535M

[HBr] = 0.423M

Explanation

H2(g) + Br2(g) <--------> 2HBr(g) K_c = 6.25 ×10¹

K_c = [HBr]²/([H2][Br2])

Initial concentration

[H2] = 1.5 ×10^-2

[Br2] = 1.5 ×10^-2

[HBr] = 5.0 ×10^-1

change in concentration

[H2] = -x

[Br2] = -x

[HBr] = +x

equilibrium concentration

[H2] = 1.5 ×10^-2 - x

[Br2] = 1.5 ×10^-2 - x

[HBr] = 5.0 ×10^-1 + 2x

so,

( 5.0 ×10^-1 + 2x)²/( 1.5×10^-2 - x)² = 6.25×10¹

5.0 ×10^-1 + 2x / 1.5 ×10^-2 - x = 7.906

0.50 + 2x = 0.1186 - 7.906x

9.906x = - 0.3814

x = 0.03850

Therefore, at equilibrium

[H2] = 0.015 - (- 0.03850 ) = 0.0535M

[Br2] = 0.015 - ( - 0.03850) = 0.0535M

[HBr] = 0.50 + ( 2× -0.03850) = 0.423M

b)

Standard free energy change of the reaction(∆G°) is related to K_c  by the following equation

∆G° = - RTlnK_c

R = gas constant, 0.008314kJ/(mol K)

T = Temperature, 308.15K

subtituting the values

∆G° = - 0.008314kJ/(mol K) × 308.15K × ln(6.25 ×10¹)

∆G°= - 0.008314kJ/(mol K) × 308.15K × 2.303log(6.25 × 10¹)

∆G° = - 10.60 kJ/mol