Calculate the standard enthalpy change for the reaction
2A+B⇌2C+2D
where the heats of formation are given in the following table:
Substance | ΔH∘f (kJ/mol) |
A | -275 |
B | -413 |
C | 223 |
D | -521 |
Express your answer in kilojoules.
Given:
Hof(A) = -275.0 KJ/mol
Hof(B) = -413.0 KJ/mol
Hof(C) = 223.0 KJ/mol
Hof(D) = -521.0 KJ/mol
Balanced chemical equation is:
2 A + B ---> 2 C + 2 D
ΔHo rxn = 2*Hof(C) + 2*Hof(D) - 2*Hof( A) - 1*Hof(B)
ΔHo rxn = 2*(223.0) + 2*(-521.0) - 2*(-275.0) - 1*(-413.0)
ΔHo rxn = 367 KJ
Answer: 367 KJ
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