Mark out the enthalpy of formation of carbon monoxide ( CO ) . Given, C s + O 2 g → CO 2 g , ∆ H = - 393.3 kJ / mol and CO g + 1 2 O 2 g → CO 2 g ; ΔH = - 282 .2 kJ / mol .

Question

Mark out the enthalpy of formation of carbon monoxide ( CO ) . Given, C s + O 2 g → CO 2 g , ∆ H = - 393.3 kJ / mol and CO g + 1 2 O 2 g → CO 2 g ; ΔH = - 282 .2 kJ / mol ., - 111.1 KJ mol -1, 676.1 KJ mol -1, - 282.8 KJ mol -1, 300.0 KJ mol -1

MARKS App · Accepted Answer

To find: C s + 1 2 O 2 g → CO g = 0 ; Δ H f ∘ = ? ? ( I ) Given: C s + O 2 g → CO 2 g ; Δ H f ∘ 1 = - 393 . 3 kJ / mol II CO s + 1 2 O 2 g → CO 2 g ; Δ H f ∘ 2 = - 282 . 2 kJ / mol III Equation (I) can be founded by subtracting equation (III) from (II), So, Δ H f ∘ = Δ H f ∘ 1 - Δ H f ∘ 2 Δ H f ∘ = - 393 . 3 - - 282 . 2 Δ H f ∘ = - 111 . 1 KJ / mol - 1

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