Consider the reaction of methane with water to create hydrogen fuel, CH_4(g) + H

Question

Consider the reaction of methane with water to create hydrogen fuel, CH_4(g) + H_2O(g) CO(g) + 3H_2(g) A reaction is carried out such that the initial concentrations of reactants are [CH__4]_0=0.2275 M and [H_2O]_0=0.1295 M, and the equilibrium concentration of [H_2]_eq is 0.2625 M. Use the information given to fill out the ICE table above. Each box must be completed. Determine the equilibrium concentration of each component of the reaction. Using the values from part b, determine a value for the equilibrium constant, K, for this reaction. If you could not solve for part b, use 0.1 M for each equilibrium concentration.

Explanation / Answer

Initial : CH4= 0.2275, [H2O] =0.1295 M [CO] =0 and [H2]=0

Let moles of H2O reacted= x

Moles of CO formed at equilibrium =x and H2= 3x

Concentration of H2,   3x= 0.2625M, x =0.2625/3=0.0875

At Equilibrium [CO] =0.0875 [H2]=0.2625 [CH4] =0.2275-0.0875=0.14 [H2O] =0.1295-0.0875=0.042

K= Equilibrium constant= [CO] [H2]3/ [CH4][H2O] = 0.0875*(0.2625)3/ (0.14*0.042)=0.2691

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