Question

For the following reaction:
2A + 2B + 3C → 3D + 2E Δ Hrxn = 999. kJ

And given that:

D → C Δ H2 = 744. kJ
A + 2C → 2D + E Δ H3 = 275. kJ

Calculate the enthalpy of the following reaction:
A + 2B → E

Answer 1

Answer:

ANSWER

Solution:- (D) +15

A+2B⟶3CΔH=?

Given:-

2A+B⟶C+2DΔH

1

=10.....(1)

A+2C⟶2D+BΔH

2

=−5

B+2D⟶A+2CΔH

3

=−ΔH

2

=5.....(2)

Adding eq

n

(1)&(2), we have

2A+B+B+2D⟶C+2D+A+2CΔH=10+5

A+2B⟶3CΔH=15

Hence the ΔH for the given reaction is +15.

Answer 2

Answer:

1468 KJ

complete answer is mentioned in the attached image