Question

Calculate the enthalpy of formation of carbon monoxide (CO) from the following data:
(i) C (s) + O2 (g) → CO2 (g) ; Δ°=− 393.5 kJ mol−1
(ii) CO (s) + 12 O2 (g) → CO2 (g); Δ°=− 283.0 kJ mol−1

Answer 1

Answer:

∆H° = –678.5 kJ.

Explanation:

Hope it will help you

Answer 2

Answer: To calculate the enthalpy of formation of carbon monoxide (CO), we can use Hess's law and the given data.

Explanation:

To calculate the enthalpy of formation of carbon monoxide (CO), we can use Hess's law and the given data. Hess's law states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.

The enthalpy of formation of a compound is the enthalpy change when one mole of the compound is formed from its constituent elements in their standard states at a pressure of 1 bar and a temperature of 298 K.

We can write the reaction for the enthalpy of formation of CO as follows:

C (s) + 1/2 O2 (g) → CO (g)

We can obtain this equation by subtracting equation (ii) from equation (i), and then multiplying equation (ii) by 1/2:

C (s) + 1/2 O2 (g) → CO (g); Δ° = ?

(i) C (s) + O2 (g) → CO2 (g); Δ°=− 393.5 kJ mol−1

(ii) 1/2 CO (s) + 6 O2 (g) → CO2 (g); Δ°=− 283.0 kJ mol−1

Now, we can apply Hess's law by adding the enthalpy changes of the two equations to obtain the enthalpy change of the desired equation:

Δ°f[CO (g)] = Δ°f[CO2 (g)] - Δ°[equation (i)] + 1/2 Δ°[equation (ii)]

Δ°f[CO (g)] = [-393.5 kJ mol-1] - [-283.0 kJ mol-1] / 2

Δ°f[CO (g)] = -110.25 kJ mol-1

Therefore, the enthalpy of formation of carbon monoxide (CO) is -110.25 kJ mol-1.

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