Question

A) calculate the standard enthalpy change for the reaction

2A+B⇌2C+2D

Use the following data:

Substance	ΔH∘f (kJ/mol)
A	-245
B	-387
C	223
D	-523

B)

For the reaction given in Part A, how much heat is absorbed when 3.70 mol of A reacts?

Answer 1

Solution :-

A)Balanced reaction equation

2A + B ----- > 2C + 2D

Formula to calculate the enthalpy change of the reaction is as follows

Delta H reaction = sum of delta Hf product - sum of Delta Hf reactant

                               = [(C*2)+(D*2)] – [ (A*2)+(B*1)]

                               = [(223*2)+(-523*2)] – [ (-245*2)+(-387*1)]

                               = 277 kJ

Therefore the enthalpy change for the reaction is 277 kJ

B) 3.70 mol A

How much heat is absorbed

Form the balanced equation and the enthalpy change value we know that

2 mol A = 277 kJ

Therefore

3.70 mol A * 277 kJ / 2 mol A = 512.5 kJ

So the 3.70 mol A absorbs 512.5 kJ heat