Question

C (s)+H2O (g) ⇌ CO (g)+ H2 (g)

Suppose 0.100 mol of H2O and solid carbon, together with 7.16 atm. of CO and 8.17 atm. of H2, are placed in a 1.00 dm3 reactor at 800°C. The value of Kp is 14.1 under these conditions.

Use the ideal gas law to determine the partial pressures of all the species at the start, and then determine the concentrations of all the species at equilibrium.

Answer 1

Solution.

$p(H2O ) = \frac{n \times R \times T}{V}$

p(H2O) = 8.80 atm.

$p(sum) = p(H2) + p(CO) + p(H2O)$

p(sum) = 24.13 atm.

$n = \frac{p \times V}{R \times T}$

$\eta(i) = \frac{n(i)}{n(sum)}$

$p(part \ i) = \eta(i) \times p(sum)$

p(part H2) = 7.96 atm.

p(part CO) = 7.24 atm.

p(part H2O) = 8.93 atm.

$Kp = \frac{p(CO) \times p(H2)}{p(H2O)}$

$Kp = \frac{(7.16+x)\times(8.17 + x)}{(8.80-x)}$

x = 2.08 atm.

p(equil H2) = 10.25 atm.

p(equil CO) = 9.24 atm.

p(equil H2O) = 6.72 atm.

$p = CRT$

$C = \frac{p}{R \times T}$

$C(equil \ H2) = 1.15*10^{-3} \ M$

$C(equil \ CO) = 1.10*10^{-3} \ M$

$C(equil \ H2O) = 7.54*10^{-4} \ M$

Answer:

p(part H2) = 7.96 atm.

p(part CO) = 7.24 atm.

p(part H2O) = 8.93 atm.

$C(equil \ H2) = 1.15*10^{-3} \ M$

$C(equil \ CO) = 1.10*10^{-3} \ M$

$C(equil \ H2O) = 7.54*10^{-4} \ M$