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Given the reactions

A.) N2(g)+O2(g)—->2NO(g) Delta H= -180.5
B.) N2(g) + 3H2(g) ——> 2NH3(g) Delta H= -91.8
C.)2H2(g)+ O2(g) —-> 2H2O(g) Delta H= -486.6

What is the delta H for the reaction below
4NH3(g)+5O2(g)—->4NO(g)+6H2O(g)

User Dariober
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1 Answer

1 vote

Answer:

The answer to your question is: ΔH = 1637.8

Step-by-step explanation:

Hess' law: This law states that the enthalpy change can be calculated even if it is not calculated directly.

"if a chemical change takes place by several routes, the overall enthalpy change is the same regardless the route".

Process

A) N2(g)+O2(g)—->2NO(g) Δ H= -180.5

B) N2(g) + 3H2(g) ——> 2NH3(g) Δ H= -91.8

C)2H2(g)+ O2(g) —-> 2H2O(g) Δ H= -486.6

The result must be:

4NH3(g)+5O2(g)—->4NO(g)+6H2O(g)

Turn letter B and multiply it by 2

4NH3 ⇒ 2N2 + 6H2 ΔH = 183.6

Multiply letter A by 2

2N2 + 2O2 ⇒ 4 NO ΔH = -361

Multiply letter C by 3

6H2 + 3O2 ⇒ 6H2O ΔH = -1459.8

Finally we add the equations up and simplify then:

4NH3 + 5O2 ⇒ 4NO + 6 H2O

And we add the ΔH = 183.6 - 361 - 1459.8

= -1637.8

User Eziz Durdyyev
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8.4k points