Δ H°comb of carbon is -x kJ mol-1. The standard formation of enthalpy of CO2(g) will be

Question

Δ H°comb of carbon is -x kJ mol-1. The standard formation of enthalpy of CO2(g) will be (A) -x kJ mol-1 (B) +x kJ mol-1 (C) -x/3 kJ mol-1 (D) Data is insufficient to predict it

Tardigrade Admin · Accepted Answer

C(s)+O2(g)→ CO2(g) Δ H°comb=-x kJmol -x=Δ H°(O2)-(0+0) Δ H°(O2)=-x kJ/mol

Q. $Δ H_{co mb}^{\circ}$ of carbon is $- x k J m o l^{- 1}$ . The standard formation of enthalpy of $C O_{2} (g)$ will be

$- x k J m o l^{- 1}$

$+ x k J m o l^{- 1}$

$- x /3 k J m o l^{- 1}$

Data is insufficient to predict it

$C (s) + O_{2} (g) \to C O_{2} (g)$
$Δ H_{co mb}^{\circ} = - x k J m o l$
$- x = Δ H^{\circ} (O_{2}) - (0 + 0)$
$Δ H^{\circ} (O_{2}) = - x k J / m o l$

Q. ΔHcomb∘​ of carbon is−xkJmol−1. The standard formation of enthalpy of CO2​(g) will be

−xkJmol−1

+xkJmol−1

−x/3kJmol−1