Answer to Question #97442 in General Chemistry for Sarah

Question

In a constant pressure calorimeter, (1.9290x10^2) mL of a 1.00 M metal chloride solution (MCl2(aq)) were added to (1.9290x10^2) mL of a 0.500 M potassium sulfate solution. The total heat capacity of the calorimeter (including all its contents) was 1.75 kJ oC-1. The temperature in the calorimeter rose from 24.50oC to (2.96x10^1) oC. Calculate the enthalpy change (in kJ/mol) for the precipitation reaction: M2+(aq) + SO42-(aq) → MSO4(s).

Answer 1

Solution.

Assume that the heat capacity of salt solutions is equal to the heat capacity of water. Now let's find a substance in excess (the limiting reagent):

$n(MCl2) = 0.1929 \times 1 = 0.1929 \ mol$

$n(K2SO4) = 0.1929 \times 0.5 = 0.0965 \ mol$

K2SO4 is limiting reagent.

According to the reaction equation, the amount of substance sulfate ions is equal to the amount of substance insoluble compound, so n(K2SO4) = n (MSO4).

According to the condition of the problem, the water (solutions) and the calorimeter were in equilibrium, then:

$Q(cal) = Q(H2O)$

$Q(cal) = C(cal) * \Delta t$

$Q(cal) = 1.75 \times 10^3 \times (29.6-24.5) = 8.925 \times 10^3 \ J$

The amount of heat is equal to the enthalpy of the reaction taken with the inverse sign, so:

$\Delta H = -Q$

$\Delta H(mole) = \frac{-Q}{n(MSO4)}$

$\Delta H(mole) = \frac{-8.925 \times 10^3}{0.0965} = -92.49 \ \frac{kJ}{mol}$

Answer:

$\Delta H(mole) = -92.49 \ \frac{kJ}{mol}$