Question

Consider the following reaction: CH4(g) + 2 H2S(g) ↔ CS2(g) + 4 H2(g) A reaction mixture initially contains 0.50 M CH4 and 0.75 M H2S. If the equilibrium concentration of H2 is 0.44 M, find the equilibrium constant (Kc) for the reaction. Consider the following reaction: CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g) A reaction mixture initially contains 0.50 M CH4 and 0.75 M H2S. If the equilibrium concentration of H2 is 0.44 M, find the equilibrium constant (Kc) for the reaction. 0.23 0.038 2.9 0.34 10.

Answer 1

Answer:

                     CH4    +      2H2S      ↔  CS2  +         4H2

initial:            0.50            0.75                 0                  0

change:        -(0.44/4)     -(0.44/2)        +(0.44/4)      +0.44

Equilibrium:    0.39             0.53              0.11            0.44

Kc = {[CS2][H2]^4}/{[CH4][H2S]^2} = 0.038

Explanation:

Initially we have reactants but no product, if 0.44 M of one of the product is found at equilibrium we can find how much was produced and how much reactants were used. when the product is formed, the reactant is being used thus the negative sign on the change. using their respective stoichiometry (their corresponding number of mole). At equilibrium the reactant used will be subtracted from the one initially present.

with equilibrium constant formula and the equilibrium concentrations, find the Kc which is 0.038

Answer 2

Answer : The value of equilibrium constant for the reaction is 0.038

Solution :

The given equilibrium reaction is,

                        $CH_4(g)+2H_2S(g)\rightleftharpoons CS_2(g)+4H_2(g)$

Initially             0.50      0.75                 0           0

At eqm.        (0.50-x)  (0.75-2x)             x          4x

The concentration of $H_2$ at equilibrium = 0.44 M

That means,

4x = 0.44

x = 0.44/4 = 0.11

The concentration of $CH_4$ at equilibrium = (0.50-x) = (0.50-0.11) = 0.39 M

The concentration of $H_2S$ at equilibrium = (0.75-2x) = (0.75-2(0.11)) = 0.53 M

The concentration of $CS_2$ at equilibrium = x = 0.11 M

The expression of $K_c$ will be,

$K_c=\frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2}$

$K_c=\frac{(0.11)\times (0.44)^4}{(0.39)\times (0.53)^2}$

$K_c=0.038$

Therefore, the value of equilibrium constant for the reaction is 0.038