Question

Generic solid, X, has a molar mass of 64.6 g/mol. In a constant-pressure calorimeter, 37.8 of X is dissolved in 263 g of water at 23.00 °C.X(s) — X(aq) The temperature of the resulting solution rises to 24.40 °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 = _________ kJ What is the enthalpy of the reaction? DeltaH rxn = ____________ kJ/mol

Answer 1

In order to find letter A, we need to calculate all this information using this calorimetry formula:

Q = mcΔT

Where:

Q = heat absorbed or released

m = mass

c = specific heat

ΔT = variation in temperature

Now placing the values of our question in the formula:

Q = 263g * (4.184J/g°C) * (24.40 - 23)

Q = 1540.54J

Q = 1.54 kJ

Now in letter B we need to use the

ΔH = Q/n

Where:

ΔH = enthalpy of the reaction

Q = the heat

n = number of moles

We need to find the number of moles first, we know that the molar mass is 64.6g/mol and we have 37.8g, therefore:

64.6g = 1 mol

37.8g = x moles

x = 0.58 moles of X

Now our equation:

ΔH = 1.54/0.58 moles

ΔH = 2.65 kJ/mol