Find the enthalpy of reaction for : 2 6 2 2 2 7 C H O 2CO 3H O 2 If the enthalpy of formation of C H ,CO and H O 2 6 2 2 are -20kJ/mol, 90kJ/mol and -70kJ/mol
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Answer:
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Explanation:
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Answer:
C
6
H
6(ℓ)
+152O
2(g)
⟶6CO
2(g)
+3H
2
O
(ℓ)
ΔH
o
Here, △H
o
=∑△
f
H
o
(products)−∑△
f
H
o
(reactants)
∴ −3267=6×△
f
H
o
(CO
2
(g))+3×△
f
H
o
(H
2
O(l))−
2
15
×△
f
H
o
(O
2
(g))−△
f
H
o
(C
6
H
6
(l))
∴−3267=6×(−393.5)+3×(−285.8)−△
f
H
o
(C
6
H
6
(l))
∴△
f
H
o
(C
6
H
6
(l)=48.6KJ/mol
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