Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 13.3 Equilibrium constant calculations-Study Notes - New Syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 13.3 Equilibrium constant calculations- Study Notes- New syllabus
Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 4 – 13.3 Equilibrium constant calculations- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.
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Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics
13.3 Calculating \( \mathrm{K_c} \) and \( \mathrm{K_p} \) from Experimental Data
The values of equilibrium constants \( \mathrm{K_c} \) and \( \mathrm{K_p} \) can be calculated using experimental equilibrium data such as concentrations or partial pressures. These calculations require correct substitution into the equilibrium expression and careful handling of units.
Calculation of \( \mathrm{K_c} \)
For a general reaction:
\( \mathrm{aA + bB \rightleftharpoons cC + dD} \)
\( \mathrm{K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}} \)
Method
- Determine equilibrium concentrations (in \( \mathrm{mol\ dm^{-3}} \)).
- Substitute values into the \( \mathrm{K_c} \) expression.
- Apply powers from coefficients.
- Include units where required.
Units of \( \mathrm{K_c} \)
- Units depend on the overall powers of concentration.
- Example: \( \mathrm{mol^{x}\ dm^{-3x}} \) where \( x = \text{(moles of products − reactants)} \).
Calculation of \( \mathrm{K_p} \)
\( \mathrm{K_p = \frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b}} \)
Method
- Determine equilibrium partial pressures (in atm).
- Substitute into the expression.
- Apply powers correctly.
Units of \( \mathrm{K_p} \)
- Units depend on powers of pressure (atm).
- Often expressed as \( \mathrm{atm^x} \).
Heterogeneous Systems
- Solids and liquids are excluded.
- Only include gaseous or aqueous species.
- Units are adjusted accordingly.
Key Features
- Use equilibrium values, not initial values.
- Apply correct powers from balanced equation.
- Exclude solids and liquids.
- Include correct units where appropriate.
Example 1
Kc Calculation:
For the reaction:
\( \mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \)
At equilibrium:
- \( \mathrm{[N_2] = 0.50} \), \( \mathrm{[H_2] = 1.50} \), \( \mathrm{[NH_3] = 0.80\ mol\ dm^{-3}} \)
▶️ Answer/Explanation
\( \mathrm{K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}} \)
\( \mathrm{K_c = \frac{(0.80)^2}{(0.50)(1.50)^3}} \)
\( \mathrm{K_c = \frac{0.64}{0.50 \times 3.375} = \frac{0.64}{1.6875} \approx 0.38} \)
Units: \( \mathrm{mol^{-2}\ dm^{6}} \) (from powers).
Example 2
Kp Calculation:
For the reaction:
\( \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \)
At equilibrium:
- \( \mathrm{p_{SO_2} = 2.0\ atm,\ p_{O_2} = 1.5\ atm,\ p_{SO_3} = 3.0\ atm} \)
▶️ Answer/Explanation
\( \mathrm{K_p = \frac{(p_{SO_3})^2}{(p_{SO_2})^2(p_{O_2})}} \)
\( \mathrm{K_p = \frac{(3.0)^2}{(2.0)^2(1.5)}} \)
\( \mathrm{K_p = \frac{9}{4 \times 1.5} = \frac{9}{6} = 1.5} \)
Units: \( \mathrm{atm^{-1}} \) (since moles decrease by 1).