Answer:
The correct option is B)
Combustion of benzene:
C
6
H
6
(l)+
2
15
O
2
(g)⟶6CO
2
(g)+3H
2
O(l)
ΔH=ΔU+PΔV; Δn=6−
2
15
=−1.5
ΔH−ΔU=PΔV
ΔH−ΔU=ΔnRT
ΔH−ΔU=−1.5×R×(127+273)
ΔH−ΔU=−600R.
Explanation:
Author:
audicampbell
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4Answer:
-26.7kJ heat is produced.
Explanation:
The combustion of [tex]CH_{4}[/tex] is given as following:
[tex]CH_{4} (g) + 2O_{2} (g)[/tex] → [tex]CO_{2} (g) + 2H_{2}O (l)[/tex]
We find the difference between the number of gaseous molecules of products and the number of gaseous molecule of reactants in the reaction.
Δ[tex]n_{g}[/tex] = [tex]n_{P} - n_{R}[/tex]
Δ [tex]n_{g}[/tex] = 1 - (1 + 2)
Δ[tex]n_{g}[/tex] = -2
We use the enthalpy equation below to find the heat.
Δ[tex]H[/tex] = Δ[tex]U +[/tex] Δ[tex]n_{g}RT[/tex]
We have the following data:
ΔH = (-20000) + (-2)(8.31)(400.15)
ΔH = -20000 + (-6650.49)
ΔH = -26650.49 J ≈ -26700 J OR -26.7kJ
Author:
elizad8fr
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