# Kc & Kp ## Kc & Kp Expressions - The **equilibrium expression** links the **equilibrium constant** to the concentrations or partial pressures of reactants and products at equilibrium - Takes into account the **stoichiometry** of the equation ### General reaction and Kc - For the general reaction: $$aA + bB \rightleftharpoons cC + dD$$

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

- Products go on top, reactants on bottom - Each concentration is raised to the power of its coefficient in the balanced equation - Square brackets $[ ]$ represent concentration in $\text{mol dm}^{-3}$ ### Gaseous Equilibria and Kp - Many industrial processes involve reactions between gases - Important for controlling reaction conditions and maximizing yields - For gaseous equilibria, we can write a different equilibrium expression using **partial pressures** instead of concentrations #### For gases only - If all substances are gases in the general reaction: $$aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$$

$K_p = \frac{p^c C(\text{eqm}) \times p^d D(\text{eqm})}{p^a A(\text{eqm}) \times p^b B(\text{eqm})}$

Where: - $p^a A$ and $p^b B$ = equilibrium reactant partial pressures (kPa) - $p^c C$ and $p^d D$ = equilibrium product partial pressures (kPa) - $a, b, c$ and $d$ = coefficients in the balanced equation

Solids and liquids are ignored in $K_p$ equilibrium expressions

$K_p$ of a reaction is constant and only changes if temperature changes

Do not use square brackets for $K_p$ expressions as these represent concentration and imply $K_c$

### Writing Kp expressions | Example Reaction | $K_p$ Expression | Units | |---|---|---| | $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ | $K_p = \frac{p^2\text{NH}_3}{p\text{N}_2 \times p^3\text{H}_2}$ | $\frac{\text{kPa}^2}{\text{kPa} \times \text{kPa}^3} = \text{kPa}^{-2}$ | | $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ | $K_p = \frac{p^2\text{SO}_3}{p^2\text{SO}_2 \times p\text{O}_2}$ | $\frac{\text{kPa}^2}{\text{kPa}^2 \times \text{kPa}} = \text{kPa}^{-1}$ | ## Kc & Kp Calculations ### Kp calculations - step-by-step process 1. Find the **mole fractions** of the gases present 2. Calculate their **partial pressures** 3. Write the **$K_p$ expression** 4. Calculate the **value of $K_p$** #### Worked example: Finding Kp **Question:** Hydrogen and bromine were mixed in a flask in a 1:1 ratio and allowed to reach equilibrium at 450 K. When equilibrium had been achieved, the total pressure in the flask was 140 kPa and the mole fraction of bromine was 0.35. The equation for the reaction is: $$\text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2\text{HBr}(g)$$ Determine the partial pressures for each gas at equilibrium and the value of $K_p$. **Answer:** **Step 1:** Since hydrogen and bromine are in 1:1 ratio at equilibrium: - Mole fraction of $\text{Br}_2$ = mole fraction of $\text{H}_2$ = 0.35 - Mole fraction of $\text{HBr}$ = $1 - (0.35 \times 2) = 0.30$ **Step 2:** Calculate partial pressures: - $p\text{Br}_2 = p\text{H}_2 = 0.35 \times 140 = 49$ kPa - $p\text{HBr} = 0.30 \times 140 = 42$ kPa **Step 3:** Write the $K_p$ expression: $$K_p = \frac{p^2\text{HBr}(g)\text{ eqm}}{p\text{H}_2(g)\text{ eqm} \times p\text{Br}_2(g)\text{ eqm}}$$ **Step 4:** Solve for $K_p$: $$K_p = \frac{42^2}{49 \times 49} = \frac{1764}{2401} = 0.735 \text{ (no units)}$$ ### Working backwards from Kp - Another style of $K_p$ calculation involves being given the value of $K_p$ and working backwards - Useful for deducing the partial pressure of one of the gases #### Worked example: Finding partial pressure from Kp **Question:** $K_p$ for the dissociation equilibrium reaction of nitrogen monoxide is 0.0021. $$2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)$$ If pure NO is introduced into a reaction flask at an initial pressure of 100 kPa, what is the equilibrium partial pressure of nitrogen? **Answer:** **Step 1:** Deduce the equilibrium partial pressure of nitrogen Let $x$ = the equilibrium partial pressure of nitrogen | | $2\text{NO}(g)$ | $\rightleftharpoons$ | $\text{N}_2(g)$ | $+$ | $\text{O}_2(g)$ | |---|---|---|---|---|---| | Initially: | 100 kPa | | 0 kPa | | 0 kPa | | At eqm: | $(100 - 2x)$ kPa | | $x$ kPa | | $x$ kPa | **Step 2:** Write the $K_p$ expression: $$K_p = \frac{p\text{N}_2(g)\text{ eqm} \times p\text{O}_2(g)\text{ eqm}}{p^2\text{NO}(g)\text{ eqm}} = 0.0021$$ $$K_p = \frac{x^2}{(100 - 2x)^2} = 0.0021$$ **Step 3:** Take the square root: $$\frac{x}{(100 - 2x)} = 0.04583$$ $$x = 0.04583 \times (100 - 2x)$$ $$x = 4.583 - 0.09166x$$ $$1.09166x = 4.583$$ $$x = \frac{4.582}{1.09166} = 4.198$$ **Therefore:** - Partial pressure of $\text{N}_2$ = 4.2 kPa (to 2 s.f.) - Partial pressure of $\text{O}_2$ would also be 4.2 kPa - Partial pressure of NO would be $100 - (2 \times 4.2) = 91.6$ kPa

Check partial pressures add up to total pressure: $29.25 + 87.75 + 33.0 = 150$ kPa confirms calculations are correct

### Example with mole fractions **Question:** A sample of 0.25 mole of nitrogen and 0.75 mole of hydrogen were reacted together to form ammonia. The equilibrium amount of nitrogen was 0.16 mol. $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$ Calculate the mole fractions of nitrogen, hydrogen and ammonia. **Answer:** | | $\text{N}_2(g)$ | $+$ | $3\text{H}_2(g)$ | $\rightleftharpoons$ | $2\text{NH}_3(g)$ | |---|---|---|---|---|---| | $t = 0$ | 0.25 mol | | 0.75 mol | | 0.00 mol | | Change | $0.25 - 0.16 = -0.09$ | | $-3 \times 0.09 = -0.27$ | | $+2 \times 0.09 = +0.18$ | | $t = \text{eqm}$ | 0.16 mol | | $0.75 - 0.27 = 0.48$ mol | | 0.18 mol | Total number of moles = $0.16 + 0.48 + 0.18 = 0.82$ | | Mole fraction calculation | Mole fraction | |---|---|---| | $\text{N}_2$ | $\frac{0.16}{0.82}$ | $= 0.195$ | | $\text{H}_2$ | $\frac{0.48}{0.82}$ | $= 0.585$ | | $\text{NH}_3$ | $\frac{0.18}{0.82}$ | $= 0.220$ |

Check mole fractions sum to 1: $0.195 + 0.585 + 0.220 = 1$

**If total pressure = 150 kPa:** Calculate the partial pressure of each gas: - Partial pressure of $\text{N}_2(g) = 0.195 \times 150 = 29.25$ kPa - Partial pressure of $\text{H}_2(g) = 0.585 \times 150 = 87.75$ kPa - Partial pressure of $\text{NH}_3(g) = 0.220 \times 150 = 33.0$ kPa