KINDLY HELP HOMEWORK DUE ON MONDAY, PLEASE SHOW ALL WORKINGS AND CALCULATION!!!

ID: 812975 • Letter: K

Question

3. Consider the following reaction at equilibrium: 3 NO(g) <=> NO2(g) + N2O(g). AH = -150 kJ.

Will the reaction proceed to the left, to the right, or neither if:

a) N2O(g) is added?

b) NO2(g) is removed?

c) The container is compressed?

d) The reaction is heated?

e) The reaction is cooled?

4.a) Write an expression for the equilibrium constant, Kc, for the following gas phase reaction at 120oC: C5H12(g) + 8O2(g) ? 5CO2(g) + 6H2O(l).

b) If, at equilibrium, [C5H12] = 0.040 M, [O2] = 0.30 M, and [CO2 ] = 1.8 M, Calculate Kc.

c) Calculate the value of Kp at 120 oC in the above

a) N2O(g) is added?

ans= reaction proceeds to left .

b) NO2(g) is removed?

Ans= Reaction proceeds to right.

c) The container is compressed?

Ans = proceed to right (as pressure increases so reaction proceeds in direction in which no. of moles decreases)

d) The reaction is heated?

Ans = Proceed to left . (as exothermic reaction)

e) The reaction is cooled?

Ans= reaction proceeds to right

4.a) Write an expression for the equilibrium constant, Kc, for the following gas phase reaction at 120oC: C5H12(g) + 8O2(g) ? 5CO2(g) + 6H2O(l).

Ans = Kc = [CO2]5 / [C5H12][O2]8

b) If, at equilibrium, [C5H12] = 0.040 M, [O2] = 0.30 M, and [CO2 ] = 1.8 M, Calculate Kc.

Ans = Kc = 1.85 / (0.04)* (0.3)8 = 7.2*106

c) Calculate the value of Kp at 120 oC

Ans= Kp= Kc (RT)dn ,, dn= 5-8-1 =-4

Kp = 7.2*106 (0.0821*393)-4

= 6.6434

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