Chemistry help pleaseeee Consider the reaction A->2B deltaH1=enthalpy 4D+2C->4B deltaH2=enthalpy A->C+2D deltaH3=?

2017-10-22 1:06 pm

What is the change in enthlapy (deltaH3) equal to for reaction:
A->C+2D deltaH3=?

A. deltaH1+deltaH2
B. deltaH1-deltaH2
C. deltaH1+1/2deltaH2
D. deltaH1-1/2deltaH2
E. 2deltaH1+1/2deltaH2

Supposedly the correct answer is D; however, no matter how many times I do it I always get B and it is the only one that seems to make sense.

Because: enthalpy of A would be deltaH1
enthalpy of C would be -1/2deltaH2
and enthalphy of D would be -1/2deltaH2

add those together and I get delta H1-2/2deltaH2
which is just deltaH1-deltaH2
am I wrong? is my professor wrong? Please let me know

Thank you in advance

回答 (1)

micatkie

2017-10-22 4:14 pm

Given:
[1]: A → 2B …… ΔH₁
[2]: 4D + 2C → 4B …… ΔH₂

Reverse the second thermochemical equation and multiply it by 1/2, we get
[1]: A → 2B …… ΔH₁
[2]: 2B → C + 2D …… -(1/2)ΔH₂

[1] + [2], and cancel 2B on the both sides.
A → C + 2D …… ΔH₃ = ΔH₁ + [-(1/2)ΔH₂] = ΔH₁ - (1/2)ΔH₂

The answer: D. ΔH₁ - (1/2)ΔH₂

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