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−1and for ΔH∘f = -363 kJ⋅mol−1how they got that answers ????

Answer 1

Answer:

ΔU°f ≈ -361 kJ/mol

ΔH⁰f  ≈ -362 kJ/mol

Explanation:

for the reaction

4*K(s) + O₂(g) → 2*K₂O(s)

that is executed on a constant volume calorimeter , then according to the fist law of thermodynamics

ΔU=Q- W , since W=∫p dV = 0 because dV=0 ( the volume V is constant)

ΔU=Q

also since all the heat produced by the reaction is absorbed by the water and the calorimeter:

-Q= (m*cv + C)*ΔT

where

m= mass of water = 1439 g

cv= specific heat capacity of water = 1 cal/g*K = 4.186 J/g*K

C= calorimeter constant = 1849 J/K

ΔT= temperature change = 1.60 K

replacing values

Q= -(m*cv + C)*ΔT=  (1439 g*4.186 J/g*K + 1849 J/K )*1.60 K= -12596 J

then ΔU°f= ΔU / n , where n= number of moles then considering the factor of 2 in the chemical equation:

ΔU°f= ΔU / n = ΔU / n= -12596 J / [(2.72 gr/2) / 39 gr/mol ] = -361208 J/ mol 3´= -361.2 kJ/mol ≈ -361 kJ/mol

then since the reaction absorbes 1 mole of  O₂ per 2 moles of K₂O(s)

Δn = - 1 mole / 2 mol = -0.50

ΔH⁰f =ΔH = ΔU - Δn*R*T =  -361.2 kJ/mol -0.50* 8.314 J/mol*K *298.15 * 1 kJ/1000 J = -362.44 kJ/mol ≈ -362 kJ/mol