Question

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the calorimeter rose by 4.5°C. Calculate the heat of reaction per mol of H20 formed.( heat capacity of the calorimeter is 50J/°C

Answer 1

-21 kJ·mol⁻¹

Step-by-step explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

`\text{Moles of acid} = \text{0.050 dm}^(3) * \frac{\text{1.0 mol}}{\text{1 dm}^(3)} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^(3) * \frac{\text{1.0 mol}}{\text{1 dm}^(3)} = \text{0.050 mol}`

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

`\text{Mass of solution} = \text{100 dm}^(3) * \frac{\text{1.00 g}}{\text{1 dm}^(3)} = \text{100 g}`

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.