Question

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



CH₃OH(g) ⇌ 2H₂(g) + CO(g)

	ΔHf° (kJ/mol)		S° (J mol-1 K-1)
CH3OH	-201.50		239.80
H2	0.00		130.68
CO	-110.53		197.66

1. Determine ΔG°_rx (in kJ) for this reaction at 1173.15 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.

Answer 1

Answer – 1) In this one there is given, reaction - CH₃OH(g) <--->2H₂(g) + CO(g)

We need to calculate the ∆G^orxn at 1173.15 K

First we need to calculate the ∆H^orxn and ∆S^orxn

We know ΔH^o rxn is ,

ΔH^o rxn = sum of the ΔH^o product – sum of the ΔH^o reactant

               = [2*ΔH^o H₂ + ΔH^oCO ] – [ΔH^o CH₃OH (g)]

             = ( 2*0.00 + -110.53 ) - (-201.5)

                = 90.97 kJ/mol

We know ΔS^o rxn is ,

ΔS^o rxn = sum of the S^o product – sum of the S^o reactant

               = [2*S^o H₂ + S^oCO ] – [S^o CH₃OH (g)]

             = ( 2*130.68 + 197.66 ) - (239.80)

                = 219.22 J/mol.K

We know the formula,

ΔG^o_rxn = ΔH^o -T ΔS^o

        = 90970 J – 1173.15 K * 219.22 J/K

      = -166208 J

       = -166.21 kJ

2) Now we know the formula

ΔG^o = -RT ln Kp

Now plugging the values in the formula,

-166208 J = -8.314 *1173.15 * ln Kp

ln Kp = 166208 J / 8.314 *1173.15

          = 17.04

Taking antiln from both side

Kp = 2.52*10⁷