Study Notes on Chemical Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
As a system approaches equilibrium, both the forward and reverse reactions are occurring simultaneously.
At equilibrium, the rates of the forward and reverse reactions are equal.
Once equilibrium is achieved, the concentrations of each reactant and product remain constant.

In a system at equilibrium, both the forward and reverse reactions are represented with a double arrow in chemical equations.
Example:
$N2O4(g) \rightleftharpoons 2NO_2(g)$

Equilibrium Constant (Keq) can be derived from the rate laws of the reactions:
- Forward reaction:
 $N2O4(g) \overset{kf}{\rightarrow} 2NO2(g)$
- Rate law:
 $Rate = kf[N2O_4]$
- Reverse reaction:
 $2NO2(g) \overset{kr}{\rightarrow} N2O4(g)$
- Rate law:
 $Rate = kr[NO2]^2$
At equilibrium, the rates of the forward and reverse reactions are equal: $Ratef = Rater$
- Thus:
 $kf[N2O4] = kr[NO_2]^2$
Rearranging gives:
$\frac{kf}{kr} = \frac{[NO2]^2}{[N2O_4]}$
Equilibrium expression becomes:
$K{eq} = \frac{[NO2]^2}{[N2O4]}$

For the generalized reaction:
$aA + bB \rightleftharpoons cC + dD$
The equilibrium expression is:
$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
For gaseous reactions, pressure is proportional to concentration, hence equilibrium can also be expressed as:
$Kp = (PC^c)(PD^d) \div (PA^a)(P_B^b)$

From the ideal gas law, we have: $PV = nRT$
- Rearranging gives:
  $P = \frac{nRT}{V}$
Thus, the relationship between $KC$ and $Kp$ is given by:
$Kp = Kc (RT)^{\Delta n}$
where
$\Delta n = (moles ext{ of gaseous product}) - (moles ext{ of gaseous reactant})$

The concentration ratio of products to reactants remains constant at equilibrium regardless of initial concentrations of reactants and products.
This principle holds true for reactions initiated from either side; starting with components on the left or right will yield the same equilibrium concentrations.

Interpretation of the value of $K$ :
- If K >> 1: the reaction is product-favored; products are predominant at equilibrium.
- If K << 1: the reaction is reactant-favored; reactants are predominant at equilibrium.

The equilibrium constant for the reverse reaction is the reciprocal of the forward reaction's constant:
- Example:
 $Kc(N2O4 \rightleftharpoons 2NO2) = 0.212$
- Thus, for the reverse:
 $Kc(2NO2 \rightleftharpoons N2O4) = \frac{1}{0.212} \approx 4.72$
If the equilibrium equation is multiplied by a factor:
- The equilibrium constant is raised to that factor's power.
- Example: For the reaction multiplied by 2:
- $K_c = (0.212)^2$
For a net reaction made up of several steps, the equilibrium constant is the product of the constants for the individual steps.

Concentrations of solids and liquids are essentially constant.
- These can be measured by density multiplied by molar mass, remaining constant at constant temperature.
- Hence, the concentrations of solids and liquids do not appear in the equilibrium expression.
- Example:
 $Kc = [Pb^{2+}][Cl^-]^2$ for $PbCl2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$

Example Problem:
- A closed system initiates with concentrations of
 $[H2] = 1.000 \times 10^{-3} M$ , $[I2] = 2.000 \times 10^{-3} M$ at 448 °C.
After reaching equilibrium, $[HI] = 1.87 \times 10^{-3} M$ .
Calculation Steps:
- Initially:
- $[H_2] = 1.000 \times 10^{-3} M$
- $[I_2] = 2.000 \times 10^{-3} M$
- $[HI] = 0$
- Changes at equilibrium:
- $[HI]$ increases by $1.87 \times 10^{-3} M$
- Thus,
 $[H2] ext{ and } [I2] ext{ will decrease by half that amount: } -9.35 \times 10^{-4} M$
- Final concentration equations post-transaction:
- $[H_2] = 1.000 \times 10^{-3} - 9.35 \times 10^{-4} = 6.5 \times 10^{-5} M$
- $[I_2] = 2.000 \times 10^{-3} - 9.35 \times 10^{-4} = 1.065 \times 10^{-3} M$
- $[HI] = 1.87 \times 10^{-3} M$
Thus, the equilibrium constant: $Kc = \frac{[HI]^2}{[H2][I_2]}$
- Plugging values in:
 $K_c = \frac{(1.87 \times 10^{-3})^2}{(6.5 \times 10^{-5})(1.065 \times 10^{-3})} = 51$

Q provides the same ratio as the equilibrium expression for a system not at equilibrium.
To calculate Q, substitute initial concentrations of reactants and products into the equilibrium expression.
If the following conditions occur:
- If $Q = K$ : system at equilibrium.
- If $Q < K$ : reaction will shift right to form products.
- If $Q > K$ : reaction will shift left to form reactants.

States:
- "If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration of any components, the system will shift its equilibrium position to counteract the disturbance."

Essential for transforming nitrogen and hydrogen into ammonia (NH3), significantly impacting agriculture.
In the process, if hydrogen (H2) is added:
- This will consume nitrogen (N2) and lead to the production of more ammonia (NH3).
The setup can help direct the reaction to favor ammonia production by removing NH3 from the system as a liquid.

Temperature dependence on equilibrium illustrated through a color change:
- Example Reaction:
 $Co(H2O)6^{2+}(aq) + 4Cl^{-}(aq) \rightleftharpoons CoCl4^{2-}(aq) + 6H2O(l)$
- Endothermic Reaction:
- Heat is considered a reactant.
- Adding heat shifts the equilibrium right, increasing the blue concentration of $CoCl4^{2-}$ and decreasing pink $Co(H2O)_6^{2+}$ concentration.
- Removing heat shifts the equilibrium left to increase $Co(H2O)6^{2+}$ and reduce $CoCl_4^{2-}$ concentration.

Catalysts enhance both forward and reverse reactions, reducing the activation energy.
The equilibrium composition remains unchanged when a catalyst is employed, though equilibrium is achieved faster.
Catalysts do not alter the position of equilibrium, i.e., the equilibrium constant (K) stays consistent regardless of reaction speed.