In an experiment involving the reaction of methane and hydrogen sulfide, 1.00 mo

Question

In an experiment involving the reaction of methane and hydrogen sulfide, 1.00 mol of CH4 , 2.00 mol of H2S, 1.00 mol of CS2 , and 2.00 mol of H2 are mixed in a 250. mL sealed flask and react at 960. degree C. At this temperature, K c = 0.0341 (6 points) CH 4 (g) + 2 H 2 S (g) <----> CS 2 (g) + 4 H 2 (g) a. In which direction will the reaction proceed to reach equilibrium? (Show calculation) b. If [CH 4 ] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other substances?

Explanation / Answer

intial concentration of components=

CH4=1/0.25=4M

H2s=2/0.25=8M

Cs2=1/0.25=4M

H2=2/0.25=8M

Kc=0.0341

Kc= (cs2)*(H^4)/Ch4*H2s^2

so let equilibrium conversion w.r.t methane be X

so

Ch4=4-X

H2s=8-2x

Cs2=4+x

H2=8+4x

so

0.0341=(4+X)*(8+4X)^4/(4-X)*(8-2X)^2

we solve for X

=-3.4,-2.7,-1.74

So as all the x values are negative the equilibrium shifts towards left side that is towards reactant side

2)

Given concentration of Ch4=5.56 so X here is = -5.56+4=-1.56

so concentration of

H2s=8-2*(-1.56)=7.12M

Cs2=4-1.56=2.44

H2=8-4(1.56)=1.76 M

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