Given that C + O 2 longrightarrow CO 2: Δ H °=- xkJ 2 CO + O 2 longrightarrow 2 CO 2: Δ H °=-y kJ the enthalpy of formation of carbon monoxide will be

Question

Given that C + O 2 longrightarrow CO 2: Δ H °=- xkJ 2 CO + O 2 longrightarrow 2 CO 2: Δ H °=-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

Tardigrade Admin · Accepted Answer

Given C + O 2 arrow CO 2, Δ H °=- xkJ...(i) 2 CO 2 arrow 2 CO + O 2 Δ H °=+ y kJ ldots (ii) or CO 2 arrow CO +1 / 2 O 2, Δ H °=+ y / 2 kJ ldots..(iii) By adding eq. (i) and (iii), we get C + O 2+ CO 2 longrightarrow CO 2+ CO +(1/2) O 2 . C +(1/2) O 2 longrightarrow CO, Δ H °= y / 2- x =( y -2 x/-2) kJ

Q. Given that $C + O_{2} ⟶ C O_{2} : Δ H^{\circ} = - x k J$
$2 CO + O_{2} ⟶ 2 C O_{2} : Δ H^{\circ} = - y k J$
the enthalpy of formation of carbon monoxide will be

$\frac{2 x - y}{2}$

$\frac{y - 2 x}{2}$

2x - y

y - 2x

Q. Given that C+O2​⟶CO2​:ΔH∘=−xkJ 2CO+O2​⟶2CO2​:ΔH∘=−ykJ the enthalpy of formation of carbon monoxide will be

22x−y​

2y−2x​

2x - y

y - 2x

Given C+O2​→CO2​,ΔH∘=−xkJ...(i) 2CO2​→2CO+O2​ΔH∘=+ykJ… (ii) or CO2​→CO+1/2O2​,ΔH∘=+y/2kJ…..(iii) By adding eq. (i) and (iii), we get C+O2​+CO2​⟶CO2​+CO+21​O2​. C+21​O2​⟶CO, ΔH∘=y/2−x=−2y−2x​kJ

Q. Given that $C + O_{2} ⟶ C O_{2} : Δ H^{\circ} = - x k J$
$2 CO + O_{2} ⟶ 2 C O_{2} : Δ H^{\circ} = - y k J$
the enthalpy of formation of carbon monoxide will be

$\frac{2 x - y}{2}$

$\frac{y - 2 x}{2}$