Question

The general form of a chemical reaction is aA + bB = cC +dD

Where A and B are reactants in the forward direction and C and D are products in the forward direction. The lower case letters are the stoichiometric coefficients for the balanced equation. The general form of the equilibrium constant equation is then:

Keq = [C]c[D]d/[A]a[B]b

Part I: Q vs. K

Q: reaction quotient can be calculated for a reaction at any concentration values of reactants and products (using the K_c expression)

At equilibrium Q=K

When equilibrium conditions are altered (say adding or subtracting reactants or products or starting from concentrations that differ from equilibrium concentrations) then Q¹K

If Q > K the system will shift to the left (to consume products and make reactants)

If Q < K the system will shift to the right (to consume reactants and make products)

For the following problems use this model reaction:

H₂(g) + F₂(g) ßà 2HF(g) Kc = 1.0 x10² @ 25°C

If the system is at equilibrium Kc=Q= [HF]²/[H₂][F₂]= 1.0 x10²

Problem 1: 5.0 mol of HF(g), 3.0 mol of H₂(g) and 1.5 mol of F₂(g) are mixed in a 1.0 L vessel.

1. Is the system at equilibrium?
2. If the system is not at equilibrium, in which direction will the system shift (left-reactants/right-products) to re-establish equilibrium conditions?

Problem 2: 55.0 mol of HF(g), 1.0 mol of H₂(g) and 0.20 mol of F₂(g) are mixed in a 10.0 L vessel.

3. Is the system at equilibrium?
4. If the system is not at equilibrium, in which direction will the system shift (left-reactants/right-products) to re-establish equilibrium conditions?

Part II: Model chemical equilibrium for the following two calculations.

Consider the following reaction: 2NH3 à 3H2  + N2                             

Equilibrium Problem Type 1. Calculating Keq from equilibrium concentrations.

In these problems it is necessary to determine all the bottom row equilibrium concentrations (the ones in [X]) and with the stoichiometric coefficients, Keq is determined.

Calculation 1. For the ammonia equilibrium above at a certain temperature equilibrium concentrations are [N2] = .10 M, [H2] = .50 M, [NH3 ] =.057 M. What is Keq? This is the easiest of problems because you are told what all the bottom row equilibria are.

	2NH3	3H2	N2
Initial (M)
Change (M)
Equilibrium (M)	.057	.50	.10

Equilibrium Problem Type 2. Using stoichiometry to complete the array.

A complex problem provides some of the initial and some of the equilibrium concentrations and you are asked to solve for the rest of the concentrations in the array.

These problems are accomplished using stoichiometry and simple substitutions for unknowns.

Calculation 2. Given Keq = 3.8 (equilibrium mixture at the same temperature as previous example) equilibrium concentrations [N2 ] = 0.3000 M, [H2 ] = 0.2000 M and initial [NH₃] = 0.04000 M. What are the initial concentrations of H₂ and N₂? Fill in all empty boxes below.

2NH3 3H2 N2 Initial (M) .04000 ? ? Change (M) Final (M) .2000 .3000

Solve for initial [N₂] and [H₂]

Equilibrium Problem Type 3. A quadratic.

Consider the following reaction:

Fe³⁺(aq) + SCN^-(aq)    à      FeSCN²⁺(aq)    K=0.333=1/3 (3SF)

Initial: 8.00 M Fe³⁺(aq), 8.00 M SCN⁻(aq), and 0.00 M FeSCN²⁺_(aq)

	Fe3+	SCN-	FeSCN2+
Initial (M)	8.00	8.00	0.00
Change (M)
Equilibrium (M)			?

To solve this problem you will need to solve a quadratic equation.

Part III: Le Châtelier

1. Le Châtelier and temperature: How would increasing the temperature affect the following equilibria?

(a) 2H2O(g) ↔ 2H2(g) + O2(g)                                                    ∆H = 484 kJ

(b) C3H8(g) + 5O2(g) ↔ 3CO2(g) + 4H2O(g)                         ∆H = -2043 kJ

(c) CH4(g) + 2 O2(g) ↔ CO2(g) + 2 H2O(l)                            ∆H = -890.4 kJ

2. Le Châtelier and concentration:
   1. What happens to C2H6 when HCl gas is added?

C2H6_(g) + Cl2_(g) ↔ C2H5Cl_(s) + HCl_(g)

2. What happens to NH3 when H2 gas is added?

                   N2(g) + 3H2(g) ↔ 2NH3(g)

3. Le Châtelier and volume/pressure:
   1. What happens to C2H6 when the volume is increased?

C2H6_(g) + Cl2_(g) ↔ C2H5Cl_(s) + HCl_(g)

2. What happens to NH3 when the volume is decreased?

N2(g) + 3H2(g) ↔ 2NH3(g)

Answer 1

Problem 1

H_2(g) + F_2(g)       2HF_(g)

Given that , Kc = 1.0 x 10²   = 100

[HF] = 5.0 mol / 1L = 5.0 M

[H₂] = 3.0 mol / 1 L = 3.0 M

[F₂] = 1.5 mol / 1L = 1.5 M

Reaction quotient, Q_c =   [HF]² /  [H₂][F₂]

= (5.0)² / (3.0 x 1.5) = 5.56

Q_c = 5.56

i) Since, Q_c is less than K_c, this indicates that, the reaction is not at equilibrium.

ii) Q_c is less than K_c, this implies that the reaction will proceed to the right i.e. more product will be produced.