Question

I want to know the steps.

Answer 1

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$H_{f}$ of $C_{6}H_{12} O_{6}$) =-1275.0

enthalpy of combustion of oxygen(Δ$H_{f}$ of $O_{2}$) = zero

enthalpy of combustion of carbon dioxide(Δ$H_{f}$ of $CO_{2}$) = -393.5

enthalpy of combustion of water(Δ$H_{f}$ of $H_{2} O$) = -285.8

To solve :

standard enthalpy of combustion

We know;

Δ$H_{f}$  = ∈Δ$H_{f}$ (products) - ∈Δ$H_{f}$ (reactants)

$C_{6}H_{12} O_{6}$ (s) +6 $O_{2}$(g) → 6 $CO_{2}$ (g)+ 6 $H_{2} O$(l)

Δ$H_{f}$ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

Δ$H_{f}$ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

Δ$H_{f}$ = 6 (-393.5) + 6(-285.8)  - 0 + 1275

Δ$H_{f}$ = -2361 - 1714 - 0 + 1275

Δ$H_{f}$ =-2800 kJ

Therefore the standard enthalpy of combustion is -2800 kJ