The following reaction was carried out in a 2.25L reaction vessel at 1100 K: C(s

Question

The following reaction was carried out in a 2.25L reaction vessel at 1100 K: C(s)+H2O(g)CO(g)+H2(g) If during the course of the reaction, the vessel is found to contain 5.50mol of C, 13.9mol of H2O, 4.00mol of CO, and 7.80mol of H2, what is the reaction quotient

question 2:

The reaction

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

has an equilibrium constant of K = 0.154.

If 6.45mol of CH4, 4.70mol of C2H2, and 10.45mol of H2 are added to a reaction vessel with a volume of 5.40L , what net reaction will occur?

The reaction will proceed to the left to establish equilibrium. The reaction will proceed to the right to establish equilibrium. No further reaction will occur because the reaction is at equilibrium.

Explanation / Answer

1) [CO] = moles of CO/volume of vessel in litres = 4/2.25 = 1.78M

[H2] =  moles of H2/volume of vessel in litres = 7.8/2.25 = 3.47 M

[H2O] =  moles of H2O/volume of vessel in litres = 13.9/2.25 = 6.18 M

Reaction Quotient = [CO]*[H2]/[H2O] = 1.78*3.47/6.18 = 0.99945

2) [CH4] =  moles of CH4/volume of vessel in litres = 6.45/5.4 = 1.194 M

[C2H2] =  moles of C2H2/volume of vessel in litres = 4.7/5.4 = 0.87 M

[H2] =  moles of H2/volume of vessel in litres = 10.45/5.4 = 1.9352 M

Now, reaction quotient = [H2]3*[C2H2]/[CH4]2 = 2.2854

Since reaction quatient > Kc

therefore backward reaction will occur.

hence the correct option is:-

The reaction will proceed to the left to establish equilibrium.

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