The reaction 2CH4(g)C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154. If

Question

The reaction 2CH4(g)C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154. If 6.25 mol of CH4, 4.70 mol of C2H2, and 11.30 mol of H2 are added to a reaction vessel with a volume of 5.60 L , what net reaction will occur? The reaction has an equilibrium constant of = 0.154. If 6.25 of , 4.70 of , and 11.30 of are added to a reaction vessel with a volume of 5.60 , what net reaction will occur?

A - The reaction will proceed to the left to establish equilibrium.

B -The reaction will proceed to the right to establish equilibrium.

C -No further reaction will occur because the reaction is at equilibrium.

Explanation / Answer

Given :

Reaction :

2CH4(g)C2H2(g)+3H2(g)

Given :

equilibrium constant of K = 0.154. If 6.25 mol of CH4, 4.70 mol of C2H2, and 11.30 mol of H2 are added to a reaction vessel with a volume of 5.60 L

Solution:

Calculation of concentration of each species

[CH4]= 6.25 mol / 5.60 L= 1.116 M

[C2H2]=4.70 mol /5.60 L = 0.8392

[H2]= 11.30 mol /5.60 L = 2.018

Calculation of Qc

Qc = [C2H2][H2]3 / [CH4]2

Lets plug all the values in above equation

= (0.8392) (2.018)3 / (1.116)2

= 5.53

Kc for this reaction is 0.154

Since the value of Qc is greater than Kc so reaction will proceed to left to establish the equilibrium

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