For each of the following reactions: (a) Write the Equilibrium expression (b) Ca

Question

For each of the following reactions: (a) Write the Equilibrium expression (b) Calculate the Equilibrium Constant, Kc (c) Determine the DIRECTION of the reaction (d) Determine if the reaction is in Equilibrium (e) Determine the effect of INCREASING the temperature (f) Determine the effect of increasing the concentration of ONE of the reactants. (g) Determine the effect of increasing the volume of the container. (h) Determine the effect of increasing the Pressure in the container. Reaction 1: N204(g) 2NO2(g) Reaction 2: CO2(g) + H20(I) H2CO3(aq) Each reaction is allowed to come to equilibrium, and then the volume is changed as indicated. Predict the effect (right shift, left shift, or no effect) of the indicated volume change. (a) CO(g) + H2O(g) CO2(g) + H2(g) (volume is decreased) (b) PCI3(g) + CI2(g) PCI5 (g) (volume is increased) (c) CaCO3(s) CaO(s) + CO2(g) (volume is increased) The following reaction Is endothermic: NH_3 (g) + H2O(I) NH4OH (aq) (a) What is the value of the Equilibrium Constant, Kc, for this reaction. (b) What is the effect of of increasing temperature of the reaction mixture. (c) What is the effect of decreasing temperature of the reaction mixture

Explanation / Answer

1.

(a) For reaction 1,

Equilibrium expression = [NO2]^2/[N2O4]

For reaction 2,

Equilibrium expression = [H2CO3]/[H2O][CO2]

(b) For reaction 1,

dGo = 2 x 51.3 - 97.82 = 4.78 kJ

dGo = -RTlnKc

4780 = -8.314 x 298 lnKc

Kc = 0.1452

For reaction 2,

dGo = -623.1 - (-394.39 - 228.61) = -0.1 kJ

dGo = -RTlnKc

-100 = -8.314 x 298 lnKc

Kc = 1.041

(c) Direction of reaction in case of,

Reaction 1 Is towards reactant as it free energy change dGo is +ve

Reaction 2 Is towards products as it free energy change dGo is -ve

(d) The reaction is not at equilibrium in first case

The reaction is almost at equilibrium in the second case.

(e) When temperature is increased, reaction 1, equilibrium shifts towards products. More products are formed.

When temperature is increased, reaction 1, equilibrium shifts towards reactants. More reactants are formed.

(f) Increasing concentration of reactant would enhance product formation in both the cases.

(g) Decreasing colume reduces pressure and reaction 1 equilibrium shifts towards product end (higher moles).

Decreasing colume reduces pressure and reaction 1 equilibrium shifts towards reactant end (higher moles).

(h) Increasing pressure would have reverse effect to that we have seen in (g).

2. Changing volume and effect on the equilibrium state of the reaction,

(a) no effect : same number of moles on both ends

(b) left shift to higher number of moles

(c) right shift to higher number of moles

3. For the given endothermic reaction,

(a) dGo = dGoproducts - dGoreactants

             = 236.65 - (-16.4 - 237.14)

             = 490.19 kJ

dGo = -RTlnKc

490.19 x 1000 = -8.314 x 298 lnKc

Kc = 1.19 x 10^-86

(b) Increasing temperature would enhance product formation for this endothermic reaction

(c) Decreasing temperature would reverse the reaction towards reactant side.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.