Given that C + O₂ → CO₂ : ∆Hº = -x kJ
2CO + O₂ → 2CO₂ : ∆Hº = -y kJ
the enthalpy of formation of carbon monoxide will be
(a) 2x - y/2
(b) y - 2x/2
(c) 2x – y (d) y – 2x
Answers
answer : option (b) (y - 2x)/2
Given that C + O₂ → CO₂ : ∆Hº = -x kJ.......(1)
2CO + O₂ → 2CO₂ : ∆Hº = -y kJ.......(2)
we have to find enthalpy of formation of CO.
i.e., C + (1/2)O₂ ⇒CO ; ∆H = enthalpy of formation.
reverse and multiple (1/2) with equation (2) we get,
CO₂ ⇒CO + (1/2)O₂ ; ∆H = -∆H°/2 = y/2 ........(3)
now adding equations (1) and (3) we get,
C + (1/2)O₂ ⇒CO ; ∆H' = ∆H + ∆H° = y/2 + (-x)
= (y - 2x)/2
hence option (b) is correct choice.
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Answer:
option b
Explanation:
y-2x/2
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