Heat absorbed and enthalpy

Question

A generic solid, X, has a molar mass of 58.8 g/mol. In a constant‑pressure calorimeter, 18.0 g of X is dissolved in 377 g of water at 23.00 °C.

The temperature of the resulting solution rises to 27.90

 °C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there is negligible heat loss to the surroundings.

How much heat was absorbed by the solution?

q=?

What is the enthalpy of the reaction?

Δ𝐻rxn=?

Answer 1

With any question dealing with energy, always remember that for a normal chemical or physical process (excluding nuclear reactions) energy is conserved. In other words, if one thing releases energy (in the form of heat or work), something else will absorb that energy. In the question given, the water temperature is increasing. This means that the water is absorbing energy in the form of heat. The heat that caused the water to increase in temperature is released from the dissolved solid X.

Another thing to remember is that energy that is absorbed (an energy deposit) is always given a positive sign, while energy released (energy withdrawal) is given a negative sign.

The equations we will use for a constant pressure calorimetry problem are:

-q_rxn = +q_soln Note: this is only true because there is negligible heat loss to the surroundings.

q_soln = m_soln*C_{s, soln}*ΔT_soln

Additionally, when at constant pressure, q_rxn = ΔH_rxn

First, since we know all the variables for the solution except heat, q, we will solve for the heat absorbed by the solution first:

q_soln = m_soln*C_{s, soln}*ΔT_soln q_soln = (377 g + 18.0) * (4.184 J/(g·°C)) * (27.90°C - 23.00°C) = 8.10 x 10³ J

That means that 8.10 x 10³ J of energy were absorbed, heating the solution (water + solid X) up 4.90°C, and since -q_rxn = +q_soln:

-q_rxn = 8.10 x 10³ J q_rxn = -8.10 x 10³ J

Therefore, the dissolving process released -8.10 x 10³ J of energy. Note: if there was not negligible heat loss to the surroundings, we would also need to account for heat lost to the surroundings, usually by using a calorimeter constant.

Next, we will solve for the enthalpy of the dissolving process, ΔH_rxn. Though it is true that q_rxn = ΔH_rxn for a system under constant pressure, ΔH_rxn is generally given in units of kJ/mol not just J, which is what was solved for above. Therefore, we need to convert from J to kJ, then into kJ/mol.

Solve for moles of solid X:

18.0 g X (1 mol X / 58.8 g X) = 0.306 mol X

Solve for ΔH_rxn by converting from J to kJ to kJ/mol:

q_rxn = -8.10 x 10³ J (1 kJ / 1000 kJ) = -8.10 kJ

ΔH_rxn = -8.10 kJ / 0.306 mol X = -26.5 kJ/mol