Question

practice problems:

1. in a calorimetry experiment, 7.46g of potassium chloride is dissolved in 100.0ml (100g) of water at an initial temperature of 24.1°c. the final temperature of the solution is 20°c. what is the molar enthalpy of solution of potassium chloride? (17kj/mol)
2. a chemist wants to determine empirically the enthalpy change for the following reaction:

mg(s) + 2hcl (aq) → mgcl₂ (aq) + h₂ (g)
the chemist uses a coffee - cup calorimeter to react 0.50g mg ribbon with 100ml of 1.00mol/l hcl. the initial temperature of the hcl is 20.4°c. after neutralization, the highest recorded temperature is 40.7°c. calculate the enthalpy change, in kj/mol of mg, for the reaction. (-4.1x10²kj/mol)

1. what mass of lithium chloride must have dissolved if the temperature of 200.0g of water increased by 6.0°c? the molar enthalpy of solution of lithium chloride is -37kj/mol. (5.7g)

pg.309/q.31 - 40

Explanation:

Problem 1

Step1: Calculate heat lost by water

$q = mc\Delta T = 100g \times 4.18J/(g^\circ C) \times (20-24.1)^\circ C = -1713.8J$
Heat absorbed by KCl: $q_{soln} = 1713.8J = 1.7138kJ$

Step2: Find moles of KCl

Molar mass of KCl = 39.1+35.5=74.6g/mol
$n = \frac{7.46g}{74.6g/mol} = 0.1mol$

Step3: Compute molar enthalpy

$\Delta H = \frac{q_{soln}}{n} = \frac{1.7138kJ}{0.1mol} \approx 17kJ/mol$

Step1: Calculate heat gained by solution

Assume solution mass = 100g, $c=4.18J/(g^\circ C)$
$q = mc\Delta T = 100g \times 4.18J/(g^\circ C) \times (40.7-20.4)^\circ C = 8485.4J = 8.4854kJ$
Heat released by reaction: $q_{rxn} = -8.4854kJ$

Step2: Find moles of Mg

Molar mass of Mg=24.3g/mol
$n = \frac{0.50g}{24.3g/mol} \approx 0.0206mol$

Step3: Compute enthalpy change

$\Delta H = \frac{q_{rxn}}{n} = \frac{-8.4854kJ}{0.0206mol} \approx -4.1\times10^2kJ/mol$

Step1: Calculate heat gained by water

$q = mc\Delta T = 200g \times 4.18J/(g^\circ C) \times 6.0^\circ C = 5016J = 5.016kJ$
Heat released by LiCl dissolution: $q_{soln} = -5.016kJ$

Step2: Find moles of LiCl

$\Delta H = -37kJ/mol = \frac{q_{soln}}{n} \Rightarrow n = \frac{q_{soln}}{\Delta H} = \frac{-5.016kJ}{-37kJ/mol} \approx 0.1356mol$

Step3: Compute mass of LiCl

Molar mass of LiCl=6.9+35.5=42.4g/mol
$m = n \times M = 0.1356mol \times 42.4g/mol \approx 5.7g$

Answer:

17kJ/mol

Problem 2