Thermochemistry, Kinetics & Equilibrium — Science

  1. Grade 11 — Energy in Chemical Reactions
  2. Grade 11 — Reaction Rates & Kinetics
  3. Grade 11 — Chemical Equilibrium

Apply it in a new setting

Problem 1

How much heat (in J) is required to raise the temperature of 150.0 g of water from 20.0°C to 80.0°C? (Specific heat of water c = 4.18 J/g°C.)
Hints

Use q = mcΔT, where ΔT is the temperature change (final − initial).

ΔT = 80.0 − 20.0 = 60.0°C; m = 150.0 g; c = 4.18 J/g°C.

Multiply: q = 150.0 × 4.18 × 60.0.

Answer and explanation

37620 J

ΔT = 80.0 − 20.0 = 60.0°C.

q = mcΔT = 150.0 g × 4.18 J/g°C × 60.0°C.

= 37,620 J (about 37.6 kJ).

Problem 2

A 200.0 g block of copper starting at 25.0°C absorbs 3080 J of heat. What is its final temperature? (Specific heat of copper c = 0.385 J/g°C.)
Hints

Rearrange q = mcΔT to find the temperature change: ΔT = q ÷ (m × c).

ΔT = 3080 ÷ (200.0 × 0.385).

Add that ΔT to the starting temperature of 25.0°C.

Answer and explanation

65 °C

ΔT = q ÷ (mc) = 3080 ÷ (200.0 × 0.385) = 3080 ÷ 77.0 = 40.0°C.

Final temperature = initial + ΔT = 25.0 + 40.0.

= 65.0°C.

Problem 3

A 50.0 g piece of metal at 100.0°C is dropped into 100.0 g of water at 20.0°C, and the mixture settles at 22.5°C. Assuming all the heat lost by the metal is gained by the water, what is the specific heat of the metal (in J/g°C)? (c_water = 4.18 J/g°C.)
Hints

First find the heat the water gained with q = mcΔT; by energy conservation the metal lost that same amount.

Water: q = 100.0 × 4.18 × (22.5 − 20.0). The metal released this same q while cooling from 100.0°C to 22.5°C.

Solve c_metal = q ÷ (m_metal × ΔT_metal), where ΔT_metal = 100.0 − 22.5 = 77.5°C.

Answer and explanation

0.27 J/g°C

Heat gained by water: q = 100.0 × 4.18 × (22.5 − 20.0) = 100.0 × 4.18 × 2.5 = 1045 J. The metal lost the same 1045 J.

The metal cooled by ΔT = 100.0 − 22.5 = 77.5°C, so c = q ÷ (m × ΔT) = 1045 ÷ (50.0 × 77.5) = 1045 ÷ 3875.

c_metal = 0.270 J/g°C.

Problem 4

The complete combustion of methane releases 890 kJ of heat per mole of CH₄ (ΔH = −890 kJ/mol). How much heat (in kJ) is released when 40.0 g of methane burns completely? (Molar mass CH₄ = 16.05 g/mol.)
Hints

The 890 kJ is per mole, so first convert the 40.0 g of methane into moles.

Moles of CH₄ = 40.0 g ÷ 16.05 g/mol.

Multiply those moles by 890 kJ/mol to get the total heat released.

Answer and explanation

2218 kJ

Moles of CH₄ = 40.0 ÷ 16.05 = 2.49 mol.

Heat released = 2.49 mol × 890 kJ/mol.

≈ 2218 kJ released.

Problem 5

Two identical amounts of solid zinc react with the same hydrochloric acid. Sample A is one solid lump; sample B is ground into a fine powder. Which reacts faster, and why?
  1. They react at exactly the same rate, since the amount of zinc is identical
  2. Sample A (the lump), because its particles are packed together more tightly and collide harder
  3. Sample B (the powder), because greater surface area lets more particles collide with the acid each second
  4. Sample A (the lump), because a solid lump stores more thermal energy than a powder
Hints

Reaction rate depends on how often reactant particles collide — think about how much zinc surface is exposed to the acid in each case.

Grinding a solid into powder dramatically increases the surface area that touches the acid.

More exposed surface means more collisions per second, which speeds the reaction up.

Answer and explanation

Sample B (the powder), because greater surface area lets more particles collide with the acid each second

A reaction speeds up when reactant particles collide more frequently; for a solid, that means exposing more surface to the acid.

A powder has far more surface area than a single lump of the same mass, so many more zinc atoms meet the acid each second.

Therefore Sample B (the powder) reacts faster because of its greater surface area (choice C).

Problem 6

For the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), a flask at equilibrium contains [H₂] = 0.20 M, [I₂] = 0.20 M, and [HI] = 1.6 M. Calculate Keq.
Hints

Write Keq as products over reactants, each concentration raised to its coefficient: Keq = [HI]² ÷ ([H₂][I₂]).

HI has coefficient 2, so it is squared; H₂ and I₂ each have coefficient 1.

Substitute: (1.6)² ÷ (0.20 × 0.20).

Answer and explanation

64

Keq = [HI]² ÷ ([H₂][I₂]) = (1.6)² ÷ (0.20 × 0.20).

= 2.56 ÷ 0.040.

Keq = 64. Since Keq ≫ 1, the products (HI) are strongly favoured.

Problem 7

For the exothermic equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat, which single change would shift the equilibrium to the RIGHT and increase the yield of ammonia?
  1. Raising the temperature of the system
  2. Removing some N₂ gas from the system
  3. Adding a catalyst
  4. Increasing the pressure on the system
Hints

Le Châtelier's principle: the system shifts to oppose a change. Count the gas moles on each side (left 1 + 3 = 4; right 2), and remember heat behaves like a product here.

Raising temperature adds a 'product' (heat) and shifts an exothermic reaction backward; removing N₂ pulls the system left; a catalyst changes only speed, not position.

Higher pressure favours the side with fewer gas molecules — compare 4 moles of gas on the left with 2 on the right.

Answer and explanation

Increasing the pressure on the system

A catalyst only speeds up reaching equilibrium without shifting it; removing N₂ shifts the system left; and because the forward reaction is exothermic, raising the temperature also shifts it left (lowering ammonia).

The left side has 4 moles of gas and the right only 2, so increasing the pressure pushes the equilibrium toward the fewer-molecule side — the right.

The change that raises the NH₃ yield is increasing the pressure (choice D).

Browse all courses