For the equilibrium,CH₄(g)+ 2 H₂S(g)$\leftrightarrows$ CS₂(g)+ 4 H₂(g),the concentrations at equilibrium are [CH₄] = 0.3322 M,[H₂S] = 0.6644 M,[CS₂] = 0.0678 M,and [H₂] = 0.2712 M at 1400.0 K.Calculate K _c.

Question

Quizplus · Accepted Answer

Using the law of mass action, we can write the expression for the equilibrium constant as: Kc = ([CS₂]^1)[H₂]^4 / ([CH₄])([H₂S]^2) Plugging in the concentrations given, we get: Kc = (0.0678^1)(0.2712^4) / (0.3322)(0.6644^2) Kc = 2.50 x 10^-10 Therefore, the correct choice is B.

For the Equilibrium,CH4(g)+ 2 H2S(g) ⇆\leftrightarrows⇆ CS2(g)+ 4 H2(g),the Concentrations at Equilibrium Are [CH4] = 0

For the Equilibrium,CH₄(g)+ 2 H₂S(g) $\leftrightarrows$ CS₂(g)+ 4 H₂(g),the Concentrations at Equilibrium Are [CH₄] = 0