Question

5. If 25 grams of calcium carbonate are decomposed in a calorimeter that contains 175

grams of water, the water cools from 85°C to 24°C. Calculate the standard heat of reaction
for this process in kilocalories.

Answer 1

First, we need to calculate the heat lost by the water:

Q = mcΔT

where:

- m = mass of water = 175 g

- c = specific heat of water = 1 cal/g°C

- ΔT = change in temperature = (24°C - 85°C) = -61°C

Q = (175 g) x (1 cal/g°C) x (-61°C) = -10,675 cal

The negative sign indicates that heat was lost by the water.

Next, we need to calculate the moles of calcium carbonate that underwent the reaction:

n = m/M

where:

- m = mass of calcium carbonate = 25 g

- M = molar mass of calcium carbonate = 100.09 g/mol

n = 25 g / 100.09 g/mol = 0.2498 mol

Assuming that the reaction was:

CaCO3(s) → CaO(s) + CO2(g)

we can calculate the heat of reaction using the following equation:

ΔHrxn = Q / n

where:

- Q = heat lost by the water = -10,675 cal

- n = moles of calcium carbonate that underwent the reaction = 0.2498 mol

ΔHrxn = (-10,675 cal) / (0.2498 mol) = -42,720 cal/mol

To convert cal/mol to kcal/mol, we divide by 100