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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 of ) =-1275.0

enthalpy of combustion of oxygen(Δ of ) = zero

enthalpy of combustion of carbon dioxide(Δ of ) = -393.5

enthalpy of combustion of water(Δ of ) = -285.8

To solve :

standard enthalpy of combustion

We know;

Δ  = ∈Δ (products) - ∈Δ (reactants)

(s) +6 (g) → 6 (g)+ 6 (l)

Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

Δ = 6 (-393.5) + 6(-285.8)  - 0 + 1275

Δ = -2361 - 1714 - 0 + 1275

Δ =-2800 kJ

Therefore the standard enthalpy of combustion is -2800 kJ

1 is the answer to this problem