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1)Which thermochemical equation and data does not agree with the other two written?

a) 2NO (g)+ O2 (g) ->2 NO2 (g) deltaH=-169.8

b) NO (g) + 1/2 O2 (g) -> NO2 (g) delta H = -56.6

c) 4 NO2 (g) -> 4 NO (g) + 2 O2 (g) delta H = +226.4

d)all three equations are in thermochemical agreement.

I chose answer A as being incorrect because b and c look correct and the value of delta for c is 4 times the amount of b so I think there are correct. A does not fit in.

2)Calculate the value of delta H for the reaction 2 CH4 (g) -> C2H6 (g) + H2 (g) given the following thermochemical equations:

C2H2 (g) + H2 (g) -> C2H4 delta H=-175.1

C2H6(g) -> C2H4 (g) + H2 (g) delta H=+136.4

C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) delta H= -376.8

The answer choices are +65.3, +338.1, -415.5, +688.3

I chose +338.1 as the answer.

3) Consider the following of sulfur trioxide according to the reaction:

2 SO2 (g) + O2 (g) -> 2 SO3 (g) deltaH= -198

What is the value of delta H for SO3 (g) -> SO2 (g) + 1/2 O2 (g) ?

a) +99
b) -396
c) +396
d) -99
e) some other value

I chose answer a, +99 because the equation is flipped and is half of the original.

I agree with all three answers. Good work.

Thank you for checking my work!!!

Can you explain for the second questions how you got +338.1? Because per my way of doing it I get Delta H as +65.3. Below is how I did it. Please let me know if incorrect.

C2H4 + H2 --> C2H6 Delta H = -136.4 (flipped)
2CH4 --> C2H2 + 3H2 Delta H = +376.8 (flipped)
C2H2 + H2 --> C2H4 Delta H = -175.1
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2CH4 --> C2H6 + H2 Delta H = +65.3