Equilibrium Kp Flashcards

Interactive active recall cards for mole fractions, partial pressures, and Kp equilibrium constants.

Mole Fractions & Partial Pressures: read all 6 questions and answers

Define mole fraction of a gas in a mixture.
The ratio of the number of moles of a specific gas to the total number of moles of all gases in the mixture.
State the formula for the mole fraction of gas A.
Mole fraction of \(\text{A} = \frac{\text{Moles of A}}{\text{Total moles of all gases}}\).
What is the sum of the mole fractions of all gases in a mixture?
The sum of the mole fractions is always exactly 1.
Define partial pressure of a gas.
The pressure that a gas would exert if it occupied the container alone.
State the formula to calculate the partial pressure of gas A.
Partial pressure of A=Mole fraction of A × Total pressure of the gas mixture.
What is the sum of the partial pressures of all gases in a mixture?
The sum of the partial pressures is equal to the total pressure of the mixture.

Kp Expressions & Units: read all 6 questions and answers

Write the \(K_p\) expression for the general reaction: \(a\text{A}(\text{g}) + b\text{B}(\text{g}) \rightleftharpoons c\text{C}(\text{g}) + d\text{D}(\text{g})\).
\(K_p = \frac{p(\text{C})^c \times p(\text{D})^d}{p(\text{A})^a \times p(\text{B})^b}\), where \(p\) represents the partial pressure of each gas.
Why are only gaseous substances included in \(K_p\) expressions?
Only gases have partial pressures. Liquids and solids have constant concentrations and do not exert gaseous pressure.
How do you determine the units of \(K_p\)?
Substitute pressure units (e.g. kPa or atm) into the \(K_p\) expression, cancel them out, and write the remaining unit expression.
State the unit of \(K_p\) for the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) if pressure is measured in kPa.
kPa⁻² (since \(\frac{\text{kPa}^2}{\text{kPa}^4} = \text{kPa}^{-2}\)).
State the unit of \(K_p\) for the reaction: H₂(g) + I₂(g) ⇌ 2HI(g).
No units (since \(\frac{\text{kPa}^2}{\text{kPa}^2}\) cancels completely).
If a gas mixture contains 0.20 mol O₂, 0.50 mol N₂, and 0.30 mol CO₂, what is the mole fraction of O₂?
Mole fraction of \(\text{O}_2 = \frac{0.20}{0.20 + 0.50 + 0.30} = \frac{0.20}{1.00} = 0.20\).

Shifts & Factors Affecting Kp: read all 6 questions and answers

Which experimental factor is the ONLY one that changes the value of \(K_p\)?
Temperature is the only factor that changes the value of \(K_p\).
How does a change in pressure affect the value of \(K_p\)?
Changing pressure has no effect on the value of \(K_p\). The partial pressures adjust to keep \(K_p\) constant.
How does a catalyst affect the value of \(K_p\)?
A catalyst has no effect on \(K_p\). It increases the rate of both forward and reverse reactions equally, allowing equilibrium to be reached faster.
If the temperature of an exothermic reaction at equilibrium is increased, what happens to \(K_p\)?
\(K_p\) decreases because the equilibrium shifts to the left (endothermic direction) to oppose the temperature rise, decreasing products and increasing reactants.
If the temperature of an endothermic reaction at equilibrium is increased, what happens to \(K_p\)?
\(K_p\) increases because the equilibrium shifts to the right (endothermic direction) to absorb heat, increasing products and decreasing reactants.
Why does increasing total pressure of the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) shift equilibrium to the right without changing \(K_p\)?
Isothermal compression increases all partial pressures by the same factor \(f\). In \(Q_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2) \times p(\text{H}_2)^3}\), the denominator increases by \(f^4\) while the numerator increases by \(f^2\), so \(Q_p < K_p\). The system shifts forward (to the right) to restore \(Q_p = K_p\), increasing the yield of NH₃.

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