Question

For a given reaction, the heat capacity (CP) of the reactants is 50.0 J/mol K and for the products is 65.0 J/mol K. If the enthalpy of reaction is − 150.0 kJ/mol at 300K, what is the best estimate of ΔrxnH at 320K?

The reaction in the previous question is carried out adiabatically starting at 300 K. What is the final temperature after the reaction goes to completion?

Δ_rxnH₃₀₀ = -150.0 kJ/mol

C_P(reactants) = 50 J/molK

C_P(products) = 65 J/molK

T_f = ? K

Answer 1

(A) According to Kirchoff's law: Cp = ΔH / ΔT

ΔH₁ = Cp_(reactants) x ΔT

ΔH₂ = Cp_(products) x ΔT

ΔH_rxn= -150.0 KJ/mol, Cp_(reactants) = 50.0 J/mol, Cp_(products) = 65.0 J/mol, To = 300 K, T1= 320 k

ΔH_T1 = ΔH_T0 + [Cp_{(products) -} Cp_(reactants) ] x ΔT...............................................eq (1)

ΔHrxn(320K) = -150.0 + {[65-50] x (320-300)} /1000

ΔH_rxn(320K)= -149.7KJ/mol...........................................................(2)

(B) As per equation (1)

ΔH_rxn = ΔH_T0 +  [Cp_{(products) -} Cp_(reactants) ] x ΔT...............................................eq (1)

ΔH_rxn = -150.0 KJ/mol, Cp_(reactants) = 50.0 J/mol, Cp_(products) = 65.0 J/mol, To = 300 K,

ΔT = (ΔH_{rxn -} ΔH_T0) / [Cp_{(products) -} Cp_(reactants) ]

ΔT = [(-149.7) - (-150) / (65 - 50)

ΔT = 0.02

where ΔT = T_f - T_o

T_f = ΔT + T_o

T_f = 0.02 + 300 = 300.02 K.....................................................(3)

Hence there is slight change in the enthalpy (A) and temperature (B)