Answer to Question #99848 in Physical Chemistry for S

Question

A sealed container was filled with 0.300 mol H2(g), 0.400 mol I2(g), and 0.200 mol HI(g) at 870 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 870 for the reaction H2(g) + I2(g) equilibrium reaction arrow 2 HI(g).

Answer 1

Solution.

$p \times V = n \times R \times T$

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

V = 65.06 L

$C(I2) = 0.0062 \ M$

$C(H2) = 0.0046 \ M$

$C(HI) = 0.0031 \ M$

$Kp = Kc \times (R \times T)^{(- \Delta \nu)}$

$For \ this \ reaction \ \Delta \nu = 0 and \ Kp = Kc$

$Kp = \frac{[HI]^2eq}{]H2]eq \times [I2]eq}$

$870 = \frac{(0.0031+x)^2}{(0.0046-x) \times (0.0062-x)}$

x = 0.0045

$[HI] = (0.0031+2\times 0.0045) \times 65.06 = 0.79 \ mol$

$[I2] = (0.0062-0.0045) \times 65.06 = 0.11 \ mol$

$[H2] = (0.0046-0.0045) \times 65.06 = 0.0065 \ mol$

Answer:

$[HI] = 0.79 \ mol$

$[I2] = 0.11 \ mol$

$[H2] = 0.0065 \ mol$