Question

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.

Answer 1

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

Step-by-step 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

Answer 2

a) Kp 0.151 at 1000K, Kp at 298.15 is 2.15x10⁻¹⁸

b) ΔG = 100.8 kJ/mol

Step-by-step explanation:

For the reaction:

C2H6(g)⇌C2H4(g)+H2(g)

Kp is defined as:

`Kp = \frac{P_{H_(2)}P_(C_2H_4)}{P_C_2H_6}`

a) It is possible to obtain pressure of each compound using mole percent of gases, that is:

`Kp = (0.266*0.266)/(0.468)` = 0.151 at 1000K

Here, you can use:

`ln(K2)/(K1)=-(dH)/(R) ((1)/(T2) -(1)/(T1) )`

Where K is K of reaction, dH is ΔH, R is gas constant (8.314 J/molK) and T are temperatures.

Replacing:

`ln(K2)/(0.151)=-(137000J/mol)/(8.314J/molK) ((1)/(298.15) -(1)/(1000K) )`

ln K2 /15.1 = 38.8

K2 / 0.151 = 1.42x10⁻¹⁷

K2 = 2.15x10⁻¹⁸

b) ΔG of reaction at 298.15K is:

ΔG = -RT ln K

Where R is gas constant (8.314J/molK), T is temperature (298.15K), K is Kp (2.15x10⁻¹⁸)

ΔG = -8.314J/molK×298.15K ln 2.15x10⁻¹⁸

ΔG = 100.8 kJ/mol