Question

Answer the following questions using the chemical reaction and thermochemical information given below:



Br₂(g) + C₅H₈(g) ⇌ 1C₅H₆(g) + 2HBr(g)

	ΔHf° (kJ/mol)		S° (J mol-1 K-1)
C5H6	139.00		274.47
HBr	-36.29		198.70
Br2	30.9		245.5
C5H8	36.00		289.66

1. Determine ΔG°_rx (in kJ) for this reaction at 1212 K. Assume ΔH°_f and S° do not vary as a function of temperature. Report your answer to two decimal places 2. Determine the equilibrium constant for this reaction. Report your answer to three significant figures in scientific notation.

2. Determine the equilibrium constant for this reaction. Report your answer to three significant figures in scientific notation.

Answer 1

ΔG = ΔH - TΔS

First calcualte the dG for all substituents:

C5H56:

ΔG = ΔH – TΔS

ΔG = 139 – 1212 *0.27447

ΔG =-193.66

HBr:

ΔG = ΔH – TΔS

ΔG = - 36.29 – 1212 *0.19870

ΔG =-277.11

Br2:

ΔG = ΔH – TΔS

ΔG = 30.9 – 1212 *0.2455

ΔG =-266.65

C5H8:

ΔG = ΔH – TΔS

ΔG = 36.0 – 1212 *0.28966

ΔG =-315.1

Now calculate the dG for the recation:

delta G(rxn) = delta G(products) - delta G(reactants)

Br₂(g) + C₅H₈(g) ⇌ 1C₅H₆(g) + 2HBr(g)
delta G(rxn) = delta G(=-193.66 +2*-277.11 ) - delta G(-266.65 ) +(-315.1) kJ/mol

delta G(rxn) =-747.88 +581.75 kJ/mol

= -166.13 kJ/mol

= -166 kJ/mol

delta G = -RTlnK

-166000 J = (-8.31 J / mole K)(1212 K)(ln K)
16.48 =ln K
K = e^16.48= 1.4 x 10^7