Question

A sample of K(s) of mass 2.720 g undergoes combustion in a constant volume calorimeter at 298.15 K. The calorimeter constant is 1849 JK−1, and the measured temperature rise in the inner water bath containing 1439 g of water is 1.60 K.

a) Calculate ΔU∘f for K2O.

b) Calculate ΔH∘f for K2O.

the answer for ΔU∘f = -362 kJ⋅mol−1

and for ΔH∘f = -363 kJ⋅mol−1

how they got that answers ????

Answer 1

The balanced reaction for the combustion of K(s) is

2 K(s) + 1/2 O2(g) ------- > K₂O(s)

Since in the above reaction 1 mole K₂O(s) is formed from its elements in their standard state, the enthalpy and internal energy calculated for the above reaction will also be equal to the standard enthalpy of formation (?H_f⁰) and standard internal energy of formation( ?U_f⁰)

Given the mass of K(s) undergoing combustion = 2.720 g

Hence moles of K(s) = mass / atomic mass = 2.720 g / 39.10 gmol^-1 = 0.06957 mol

Here the rise in temperature conforms that heat is liberated during the reaction. Hence ?H_f⁰ is negative.

Given heat capacity of calorimeter, C = 1849 JK^-1

mass of water, m = 1439 g

rise in temperature, dT = 1.60 K

Hence heat absorbed by water, Q1 = mxsxdT = 1439 g x 4.184J.g^-1.K^-1x1.60K = 9630 J

Heat absorbed by calorimeter, Q2 = CxdT = 1849 JK^-1x1.60K = 2960 J

Hence total heat released during the burning of 0.06957 mol of K(s) = - (Q1+Q2)

= - (9630 J+2960 J) = - 12600 J = - 12600J x (1KJ / 1000J) = - 12.6 KJ

2 K(s) + 1/2 O2(g) --> K₂O(s)

2 mol 0.5 mol 1 mol

In the above balanced reaction 2 mol of K(s) forms 1 mol of K₂O(s).

Hence to find ?H_f⁰ ,we need to find the heat evolved during the combustion of 2 mol of K(s).

Hence heat evolved during the combustion of 2 mol of K(s)

= ?H_f⁰ = (- 12.6 KJ / 0.06957 mol of K) x (2 mol K / 1 mol K2O) = - 362 KJ / mol K2O

Hence ?H_f⁰ = - 362 KJ.mol^-1 (answer)

Since a constant volume is maintained inside the calorimeter, dV = 0

Hence ?H_f⁰ = ?U_f⁰ + PxdV

=>   ?H_f⁰ = ?U_f⁰ + Px0 = ?U_f⁰

Hence   ?H_f⁰  = ?U_f⁰ = - 362 KJ.mol^-1 (answer)

Hope this helps