The answer for the following problem is described below.
Therefore the standard enthalpy of combustion is -2800 kJ
Explanation:
Given:
enthalpy of combustion of glucose(Δ[tex]H_{f}[/tex] of [tex]C_{6}H_{12} O_{6}[/tex]) =-1275.0
enthalpy of combustion of oxygen(Δ[tex]H_{f}[/tex] of [tex]O_{2}[/tex]) = zero
enthalpy of combustion of carbon dioxide(Δ[tex]H_{f}[/tex] of [tex]CO_{2}[/tex]) = -393.5
enthalpy of combustion of water(Δ[tex]H_{f}[/tex] of [tex]H_{2} O[/tex]) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ[tex]H_{f}[/tex] = ∈Δ[tex]H_{f}[/tex] (products) - ∈Δ[tex]H_{f}[/tex] (reactants)
[tex]C_{6}H_{12} O_{6}[/tex] (s) +6 [tex]O_{2}[/tex](g) → 6 [tex]CO_{2}[/tex] (g)+ 6 [tex]H_{2} O[/tex](l)
Δ[tex]H_{f}[/tex] = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ[tex]H_{f}[/tex] = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ[tex]H_{f}[/tex] = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ[tex]H_{f}[/tex] = -2361 - 1714 - 0 + 1275
Δ[tex]H_{f}[/tex] =-2800 kJ
Therefore the standard enthalpy of combustion is -2800 kJ
