What is the electron of (Kr)5s24d10
1 answer:
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The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
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
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>