How much heat was released by the solution? & What is the enthalpy of the reaction?

Question

Potassium nitrate, KNO3, has a molar mass of 101.1 g/mol. In a constant-pressure calorimeter, 34.1 g of KNO3 is dissolved in 279 g of water at 23.00 °C.

KNO3(s)−→−−H2OK+(aq)+NO−3(aq)

The temperature of the resulting solution decreases to 20.90 °C. Assume that the resulting 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 released by the solution?

1. 𝑞soln= ? kJ

What is the enthalpy of the reaction?

1. Δ𝐻rxn= ? kJ/mol

Answer 1

q_soln = mcΔT = (34.1+279 grams)(4.184 J/gK)(-20.90 K) = Joules (neg. sign means that q is leaving solution. and entering endothermic process. Question is asked as if q is heat leaving soln, so it is positive.)

ΔH_rxn = -q_soln/n_{Pot. Nit.} if q solution has sign convention for entering solution (ΔH will turn out >0 this way) The convention for ΔH_rxn is per mole of reaction; and the potassium nitrate has a coefficient of 1 we can use it as the moles of reaction.

n_KNO3 = mass / molar mass = moles

It asks to convert Joules to kJ.

Please consider a tutor. Take care.