Question

Let A represent cyclopentene and let B represent cyclopentadiene. The reaction of interest can then be written as I2(g) + A(g) = 2HI(g) + B(g). When the reference pressure is chosen as 101325 Pa, the dimensionless equilibrium pressure constant for this reaction varies with temperature according to the equation, log10(kp)=7.55+(1844 K/T) (a) Calculate and ΔHm at 400oC. (b) Calculate the final equilibrium concentrations at 400oC of I2, A, HI, and B if we start with 0.100 M I2 and 0.100 M A.

Answer 1

I₂ (g) + A (g) --> 2 HI (g) + B (g)

Log10(Kp) = 7.55 + 1844 / T

T = 400 ⁰C = 673.15 K

R = 8.314 J/mol-K

Ln(Kp) / 2.3026 = 7.55 + 1844 / T

Ln(Kp) = 17.38 + 4245.97 / T … eqn(1)

a)

According to vant hoff equation,

d ln(Keq) / dT = ΔH / (RT²) .. eqn(2)

For gases, ln (Keq) = ln(Kp)

d ln(Kp) / dT = -4245.97 / T²    … eqn(3) (taking derivative of eqn (1))

Comparing equation (2) & (3) we get,

ΔH / R = -4245.97

ΔH = -4245.97 * 8.314 = - 35301 J/mol

b)

Kp at 400 C:

Log(Kp) = 7.55 + 1844 / 673.15

Log(Kp) = 10.29

Kp = 10^10.29 = 1.95 x 10¹⁰

Kc = Kp / (RT)^Δn

Change in gaseous moles, Δn = 2 + 1 – 1- 1 = 1

Kc = 1.95 x 10¹⁰ / (8.314 * 673.15)¹

= 3.48 x 10⁶

I₂ (g) + A (g) --> 2 HI (g) + B (g)

(0.1 – x) + (0.1 – x) --> 2x + x

Kc = [HI]² [B] / ([I₂] [A])

= (2x)² x / (0.1 –x )²

= 4 x³ / (0.1 –x)²

Solving we get x = 0.099967 ~ 0.1

[HI] = 2 x = 0.2 M

[B] = x = 0.1 M

[I₂] = 0.1 – x = 3.3 x 10^-5 M

[A] = 0.1 – x = 3.3 x 10^-5 M