Question

55. When 280 g of N. is converted into ammonia. Z kJ
of heat is evolved. What is the enthalpy of formation
of ammonia?
(1) 20 Z kJ
(2) -0.05 Z kJ
(3) 280 Z kJ
(4) - 28 ZkJ​

Answer 1

Answer:

The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.

Explanation:

Our target equation is

C(s) +½O₂(g) → CO(g); ΔH=?

We have the following information:

1. C(s) + O₂(g) → CO₂(g); ΔH=-393 kJ

2. 2CO(g) + O₂ → 2CO₂(g); ΔH=-588 kJ

To solve this problem, we use Hess's Law.

Our target equation has C(s) on the left hand side, so we re-write equation 1:

1. C(s) + O₂(g) → CO₂(g); ΔH=-393 kJ

Our target equation has CO(g) on the right hand side, so we reverse equation 2 and divide by 2.

3. CO₂(g) → CO(g) + ½O₂; ΔH=+294 kJ

That means that we also change the sign of ΔH and divide by 2.

Then we add equations 1 and 3 and their ΔH values.

This gives

C(s) +½O₂(g) → CO(g); ΔH=-99 kJ

Using your numbers, the standard enthalpy of formation of carbon monoxide is

-99 kJ/mol.

I think that your value for the heat of combustion of CO is incorrect. It should be

-566 kJ.

This would give the correct value of -110 kJ for the heat of formation of CO