The following reaction was carried out in a 3.25 L reaction vessel at 1100 K: C(

Question

The following reaction was carried out in a 3.25 L reaction vessel at 1100 K: C(s) + H_2O(g) CO(g) + H_2(g) If during the course of the reaction, the vessel is found to contain 8.25 mol of C, 121 mol of H_2O, 4.00mol of CO, and 7.50 mol of H_2, what is the reaction quotient Q? Enter the reaction quotient numerically. The reaction 2CH_4(g) C_2H_2(g) + 3H_2(g) has an equilibrium constant of K = 0.154. If 6.45 mol of CH 4.85 mol of C_2H_2, and 10.55 mol of H_2 are added to a reaction vessel with a volume of 5.00L, 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

Given: C (g) + H2O (g) CO (g) + H2(g)

No. of moles of C = 8.25 moles, H2O = 12.1 moles CO =4 moles H2 = 7.5 moles

Volume of vessel = 3.25 L

[C]= 8.25 mol / 3.25 L = 2.5385 mol/L (or M)

[H2O]= 12.1 mol / 3.25 L = 3.723 mol/L (or M)

[CO]= 4 mol / 3.25 L = 1.23 mol/L (or M)

[H2]= 7.5 mol / 3.25 L = 2.308 mol/L (or M)

SOLUTION

Write the Q formula:

Qc = [CO][H2] / [C][H2O]

Plug in given concentration values:

Qc=(1.23)(2.308) / (2.5385)(3.723)

Q=0.3016 Answer

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