Consider the equilibrium:

C2H6(g)⇌C2H4(g)+H2(g).

At 1000. K and constant total pressure of 1.00 bar, C2H6(g) is introduced into a reaction vessel. The total pressure is held constant at 1 bar and at equilibrium the composition of the mixture in mole percent is H2(g): 26.6 %, C2H4(g): 26.6 %, and C2H6(g): 46.8 %.

Required:

(a) Calculate Kp at 1000 K.

If ΔH of reaction = 137 kJ/mol, calculate the value of Kp at 298.15 K.

(b) Calculate ΔG of reaction for this reaction at 298.15 K.

1) Kp at 1000 K = 0.151

2) Kp at 298.15 K = 2.15*10^-18

3) ΔG = 1.0 *10^5 J/mol

Explanation:

Step 1: Data given

Temperature = 1000 K

The pressure stays cosntant at 1.00 bar

the composition of the mixture in mole percent is:

H2(g): 26.6 %

C2H4(g): 26.6 %

C2H6(g): 46.8 %.

Step 2: The balanced equation

C2H6(g)⇌C2H4(g)+H2(g)

Step 3: Calculate Kp at 1000 K.

Kx = ((pC2H4)*pH2)/ (pC2H6)

Kx = (0.266 * 0.266) / (0.468)

Kx = 0.151

Step 4: If ΔH of reaction = 137 kJ/mol, calculate the value of Kp at 298.15 K.

Kp(1000) = Kx * P/P°   since the pressure stays constant at 1 bar

Kp (1000K) = 0.151 * 1

Kp(1000K) = 0.151

ln Kp(298.15K) = ln Kp (1000K) - ΔH / R((1/298.15) - (1/1000))

ln Kp(298.15K) = -1.89 (137 *10^3 J/mol  / 8.314 J/mol*K) ((1/298.15) - (1/1000))

ln Kp(298.15K) = -40.68

Kp(298.15K) = 2.15*10^-18

Step 5: Calculate ΔG of reaction for this reaction at 298.15 K.

ΔG(298.15K) = -R*T*ln Kp(298.15K)

ΔG(298.15K) = -8.314 J/mol*K * 298.15 * -40.68

ΔG(298.15K) = 100838 K/mol = 1.0 * 10^5 J/mol

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your answers are a,b,,e and f

i hope this