Once a reversible reaction reaches equilibrium, how do the forward and reverse processes behave?
They continue at equal rates
✓
Both stop completely
Only the forward process continues
Only the reverse process continues
Correct Answer
They continue at equal rates
Equilibrium is dynamic: reactants keep turning into products and products keep turning back into reactants, at the same rate, so concentrations stop changing. Both stop completely is wrong because molecules are still reacting. Only the forward process continues is wrong because that would keep using up reactants. Only the reverse process continues is wrong because that would keep using up products.
Category: Dynamic Equilibrium
Question 2
Which is the equilibrium constant expression, Kc, for N2(g) + 3H2(g) ⇌ 2NH3(g)?
[N2][H2]^3/[NH3]^2
[NH3]^2/([N2][H2]^3)
✓
[NH3]/([N2][H2])
[NH3]^2/([N2] + [H2]^3)
Correct Answer
[NH3]^2/([N2][H2]^3)
Kc places products over reactants, with each concentration raised to the power of its coefficient in the balanced equation, so it is [NH3]^2/([N2][H2]^3). [N2][H2]^3/[NH3]^2 is upside down. [NH3]/([N2][H2]) ignores the coefficients. [NH3]^2/([N2] + [H2]^3) adds reactant terms that must be multiplied.
Category: Equilibrium Expressions
Question 3
Which is the Kc expression for CaCO3(s) ⇌ CaO(s) + CO2(g)?
[CaO][CO2]/[CaCO3]
[CaCO3]/[CaO]
[CO2]
✓
1/[CaCO3]
Correct Answer
[CO2]
Pure solids have a relative concentration of 1 and are left out of equilibrium expressions, so only the gas remains and Kc = [CO2]. [CaO][CO2]/[CaCO3] wrongly includes both solids. [CaCO3]/[CaO] contains only solids and drops the gas. 1/[CaCO3] includes a solid and omits the product gas.
Category: Equilibrium Expressions
Question 4
A reaction mixture has Q = 0.50 while K = 4.0 at the same temperature. What happens as the system moves toward equilibrium?
Net reaction runs in reverse
No net reaction occurs
The value of K falls to 0.50
Net reaction runs forward
✓
Correct Answer
Net reaction runs forward
When Q is less than K, the mixture has too little product, so the net reaction proceeds forward until Q rises to equal K. Net reaction runs in reverse happens when Q is greater than K. No net reaction occurs only when Q equals K. The value of K falls to 0.50 is wrong because K is fixed at a given temperature; Q changes, not K.
Category: Reaction Quotient
Question 5
A reaction has K = 1 × 10^30 at 25 °C. What does this value indicate?
Products are strongly favored at equilibrium
✓
Reactants are strongly favored at equilibrium
Equilibrium is reached almost instantly
Reactants and products are present in equal amounts
Correct Answer
Products are strongly favored at equilibrium
A very large K means the ratio of products to reactants at equilibrium is huge, so the reaction goes nearly to completion. Reactants are strongly favored at equilibrium describes a very small K. Equilibrium is reached almost instantly is wrong because K says nothing about speed. Reactants and products are present in equal amounts does not follow from any particular K value.
Category: Equilibrium Expressions
Question 6
At a certain temperature, K = 4.0 for A(g) ⇌ 2B(g). What is K for 2B(g) ⇌ A(g) at that temperature?
4.0
0.25
✓
2.0
16
Correct Answer
0.25
Reversing a reaction inverts its equilibrium expression, so the new constant is the reciprocal of the old one: 1/4.0 = 0.25. 4.0 is the constant for the forward direction only. 2.0 is the square root, which would apply to halving the equation. 16 is the square, which would apply to doubling the equation.
Category: Manipulating K
Question 7
Reaction 1, A ⇌ B, has K1 = 2.0, and Reaction 2, B ⇌ C, has K2 = 3.0 at the same temperature. What is K for the overall reaction A ⇌ C?
5.0
1.5
6.0
✓
0.67
Correct Answer
6.0
When two reactions are added to give an overall reaction, the equilibrium constant of the overall reaction is the product of the individual constants: 2.0 × 3.0 = 6.0. 5.0 comes from adding the constants, which is incorrect. 1.5 comes from dividing K2 by K1. 0.67 comes from dividing K1 by K2.
Category: Manipulating K
Question 8
PCl3(g) + Cl2(g) ⇌ PCl5(g) is at equilibrium in a rigid container. More Cl2 is injected at constant temperature. What is the result once equilibrium is restored?
The amount of PCl5 decreases
The value of Kc increases
The amount of PCl3 increases
The amount of PCl5 increases
✓
Correct Answer
The amount of PCl5 increases
Adding a reactant is a stress that the system relieves by using up some of the added Cl2, so the equilibrium shifts toward product and more PCl5 forms. The amount of PCl5 decreases is the opposite of the shift. The value of Kc increases is wrong because only temperature changes K. The amount of PCl3 increases is wrong because PCl3 is consumed along with Cl2 as the shift proceeds.
Category: Le Chatelier's Principle
Question 9
A product is added to a reaction mixture at equilibrium. Immediately after the addition, how does Q compare with K?
Q is greater than K
✓
Q is less than K
Q is equal to K
Q is equal to zero
Correct Answer
Q is greater than K
Adding product raises the numerator of Q, so Q becomes larger than K, and the net reaction then runs in reverse until Q falls back to K. Q is less than K would follow from adding a reactant or removing a product. Q is equal to K describes the mixture before the addition. Q is equal to zero would require that no product be present at all.
Category: Reaction Quotient
Question 10
The industrial synthesis N2(g) + 3H2(g) ⇌ 2NH3(g) is run at high pressure. Why does high pressure increase the yield of NH3?
High pressure raises the value of K
The product side has fewer moles of gas
✓
High pressure makes the reaction endothermic
High pressure lowers the activation energy
Correct Answer
The product side has fewer moles of gas
Raising the pressure by compressing the mixture shifts a gas-phase equilibrium toward the side with fewer moles of gas; here 4 moles of reactant gas form 2 moles of product gas. High pressure raises the value of K is wrong because K depends only on temperature. High pressure makes the reaction endothermic is wrong because pressure does not change the sign of the enthalpy change, and this reaction is exothermic. High pressure lowers the activation energy is wrong because that is what a catalyst does.
Category: Le Chatelier's Principle
Question 11
What does adding a catalyst do to a reaction mixture that has not yet reached equilibrium?
It raises K, so more product forms
It lowers K, so less product forms
It speeds the approach to equilibrium without changing K
✓
It shifts equilibrium toward fewer gas moles
Correct Answer
It speeds the approach to equilibrium without changing K
A catalyst lowers the activation energy of the forward and reverse reactions by the same amount, so equilibrium is reached sooner but with the same composition and the same K. It raises K, so more product forms and It lowers K, so less product forms are wrong because a catalyst does not change K. It shifts equilibrium toward fewer gas moles describes a pressure increase, not a catalyst.
Category: Le Chatelier's Principle
Question 12
For H2(g) + I2(g) ⇌ 2HI(g) at a certain temperature, an equilibrium mixture contains [H2] = 0.10 M, [I2] = 0.20 M and [HI] = 0.60 M. What is Kc, to two significant figures?
30
72
0.056
18
✓
Correct Answer
18
Kc = [HI]^2/([H2][I2]) = (0.60)^2/(0.10 × 0.20) = 0.36/0.020 = 18. 30 comes from forgetting to square [HI]. 72 comes from doubling [HI] before squaring instead of using the coefficient as an exponent. 0.056 is the reciprocal, from writing the expression upside down.
Category: Equilibrium Calculations
Question 13
For A(g) ⇌ 2B(g), a flask initially contains [A] = 1.00 M and no B. At equilibrium, [B] = 0.40 M. What is [A] at equilibrium?
0.80 M
✓
0.60 M
0.40 M
1.20 M
Correct Answer
0.80 M
Each mole of A that reacts forms 2 moles of B, so forming 0.40 M of B uses 0.20 M of A, leaving 1.00 − 0.20 = 0.80 M. 0.60 M comes from ignoring the 1:2 mole ratio. 0.40 M simply repeats the concentration of B. 1.20 M wrongly adds to A instead of subtracting.
Category: Equilibrium Calculations
Question 14
Equilibrium calculations often drop a small change x from a term such as (0.10 − x). When is this simplification usually justified?
When K is much larger than the starting concentration
When K is much smaller than the starting concentration
✓
When the reaction is carried out at high temperature
When a catalyst is added to the mixture
Correct Answer
When K is much smaller than the starting concentration
If K is very small compared with the starting concentration, so little reactant is used that x is negligible next to the initial amount; the result should still be checked. When K is much larger than the starting concentration, the reaction goes far toward products and x is not small. When the reaction is carried out at high temperature says nothing about whether x is small. When a catalyst is added to the mixture is irrelevant because a catalyst does not change K or the equilibrium amounts.
Category: Equilibrium Calculations
Question 15
For A(g) ⇌ B(g) + C(g), Kc = 1.0 × 10^-6. A flask starts with [A] = 0.010 M and no B or C. Assuming the change in [A] is negligible, what is [B] at equilibrium?
1.0 × 10^-8 M
1.0 × 10^-2 M
1.0 × 10^-4 M
✓
1.0 × 10^-3 M
Correct Answer
1.0 × 10^-4 M
With [B] = [C] = x and [A] ≈ 0.010 M, x^2/0.010 = 1.0 × 10^-6, so x^2 = 1.0 × 10^-8 and x = 1.0 × 10^-4 M, which is only 1% of 0.010 M, so the assumption holds. 1.0 × 10^-8 M is x^2, from forgetting the square root. 1.0 × 10^-2 M comes from dividing K by 0.010 before taking the root. 1.0 × 10^-3 M is the square root of K alone, ignoring the starting concentration.
Category: Equilibrium Calculations
Question 16
Which is the Kp expression for 2SO2(g) + O2(g) ⇌ 2SO3(g)?
(P_SO2)^2(P_O2)/(P_SO3)^2
2P_SO3/(2P_SO2 × P_O2)
(P_SO3)/[(P_SO2)(P_O2)]
(P_SO3)^2/[(P_SO2)^2(P_O2)]
✓
Correct Answer
(P_SO3)^2/[(P_SO2)^2(P_O2)]
Kp is written like Kc but with partial pressures: product pressures over reactant pressures, each raised to its coefficient, giving (P_SO3)^2/[(P_SO2)^2(P_O2)]. (P_SO2)^2(P_O2)/(P_SO3)^2 is inverted. 2P_SO3/(2P_SO2 × P_O2) multiplies by the coefficients instead of using them as exponents. (P_SO3)/[(P_SO2)(P_O2)] ignores the coefficients.
Category: Equilibrium Expressions
Question 17
For which gas-phase reaction does Kp equal Kc at any temperature?
H2(g) + Cl2(g) ⇌ 2HCl(g)
✓
N2(g) + 3H2(g) ⇌ 2NH3(g)
PCl5(g) ⇌ PCl3(g) + Cl2(g)
2NO2(g) ⇌ N2O4(g)
Correct Answer
H2(g) + Cl2(g) ⇌ 2HCl(g)
Kp = Kc(RT)^Δn, where Δn is moles of gaseous products minus moles of gaseous reactants, so the two are equal when Δn = 0; H2 + Cl2 gives 2 moles of gas from 2. N2(g) + 3H2(g) ⇌ 2NH3(g) has Δn = −2. PCl5(g) ⇌ PCl3(g) + Cl2(g) has Δn = +1. 2NO2(g) ⇌ N2O4(g) has Δn = −1.
Category: Equilibrium Expressions
Question 18
The reaction N2O4(g) ⇌ 2NO2(g) is endothermic. What happens when the equilibrium mixture is heated?
More N2O4 forms, and K decreases
More NO2 forms, and K increases
✓
More NO2 forms, and K is unchanged
No shift occurs, because K is fixed
Correct Answer
More NO2 forms, and K increases
For an endothermic reaction, heat acts like a reactant, so adding heat shifts the equilibrium toward NO2, and because temperature changes K, the constant becomes larger. More N2O4 forms, and K decreases describes cooling. More NO2 forms, and K is unchanged is wrong because a temperature change alters K. No shift occurs, because K is fixed is wrong because K is fixed only at constant temperature.
Category: Le Chatelier's Principle
Question 19
What is the name of the principle stating that, at a given temperature, the reaction quotient of a system at equilibrium has one constant value?
Le Chatelier's principle
Hess's law
Law of mass action
✓
Law of conservation of mass
Correct Answer
Law of mass action
The law of mass action says that when a reversible reaction is at equilibrium at a given temperature, Q takes a constant value, which is the equilibrium constant K. Le Chatelier's principle describes how an equilibrium shifts in response to a stress. Hess's law concerns adding enthalpy changes of reaction steps. Law of conservation of mass states that mass is neither created nor destroyed in a reaction.
Category: Dynamic Equilibrium
Question 20
For A(g) + B(g) ⇌ C(g), Kc = 10 at a certain temperature. A mixture contains [A] = 0.50 M, [B] = 0.20 M and [C] = 2.0 M. What is Qc?
0.050
2.9
0.20
20
✓
Correct Answer
20
Qc = [C]/([A][B]) = 2.0/(0.50 × 0.20) = 2.0/0.10 = 20; because Q is greater than K, the net reaction will run in reverse. 0.050 is the reciprocal, from inverting the expression. 2.9 comes from adding the reactant concentrations instead of multiplying them. 0.20 comes from multiplying all three concentrations together.
Category: Reaction Quotient