For the reaction given in part a. how much heat is absorbed when 3.20 mol of a reacts?

Answer

Part A Calculate the standard enthalpy change for the reaction
2A+B⇌2C+2D Use the following data: Substance ΔH∘f (kJ/mol) A -235 B
-381 C 197 D -477 Express your answer to three significant figures
and include the appropriate units.

To calculate the standard enthalpy change for the reaction uses
the following expression:

2A+B⇌2C+2D

ΔH∘f (kJ/mol) A -235, B -381 ,C 197 ,D -477

ΔH reaction = ΔH product – ΔH reactants

ΔH reaction = [2C + 2D] – [2A + B]

ΔH reaction = [2*197 + 2*-477] – [2*-235 + (-381)] kJ/mol

ΔH reaction = [394 -954] – [-470 -381] kJ/mol

ΔH reaction = + 291.00 kJ/mol

Part B For the reaction given in Part A, how much heat is
absorbed when 3.70 mol of A reacts? Express your answer to three
significant figures and include the appropriate units.

when 3.70 mol of A reacts then reaction becomes :

3.70 A+ 1.85 B⇌3.70 C+ 3.70 D

ΔH reaction = ΔH product – ΔH reactants

ΔH reaction = [3.70C + 3.70 D] – [3.70A + 1.85 *B]

ΔH reaction = [3.70*197 + 3.70*-477] – [3.70*-235 + (1.85-381)]
kJ/mol

ΔH reaction = +728.9 -1764.9 +869.5 + 704.85 kJ/mol

ΔH reaction = + 538.35 kJ/mol