Part B The following reaction was carried out in a 3.50 L reaction vessel at 110

Question

Part B The following reaction was carried out in a 3.50 L reaction vessel at 1100 K: If during the course of the reaction, the vessel is found to contain 6.00 mol of C, 15.7 mol of H2O, 3.10 mol of CO, and 6.80 mol of H2, what is the reaction quotient Q? Part C The reaction has an equilibrium constant of K = 0.154. If 6.65 mol of CH4, 4.30 mol of C2 H2, and 11.65 mol of H2 are added to a reaction vessel with a volume of 5.00 L. 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

Part B : The reaction is C(s) +H2O (g) <=>CO(g) +H2(g)

Reaction coefficient Q= [CO][H2}/ [H2O] ( since Carbon is a solid, not included)

[] stands for concentration

[CO] =3.1/3.5 =0.8857 mol/li

[H2] = 6.8/3.5 = 1.943 mol/lit

[H2O]= 15.7/3.5=4.48 mol/lit

Q= 0.8857*1.943/ 4.48 =0.3836

Part C:

The reactions is 2CH4---> C2H2+3H2

The reaction coefficient Q = [C2H2] [H2]3/ [CH4]2

[ ] indicates concentration

[C2H2] =4.3/5 =0.86M

[H2] = 11.65/ 5 =2.33 M

[CH4] = 6.65/5=1.33 M

Q = 0.86*(2.33)3/ (1.33)2 =6.14

Qc (6.14) is > Kc (0.154)

Hence, Qc has to be reduced . This is possible only the products get converted into reactants. The system procceds from right to left to establish equilibrium (First one is correct answer)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---