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.
Are these answers correct?
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
1) Your answer for question 1 is correct. The thermochemical equation and data that does not agree with the other two equations is option a). This is because the value of delta H (-169.8) for the reaction 2NO (g) + O2 (g) -> 2NO2 (g) does not match the delta H values of the other two equations.
2) Your answer for question 2 is incorrect. To calculate the value of delta H for the given reaction 2 CH4 (g) -> C2H6 (g) + H2 (g), you need to use the given thermochemical equations and perform the necessary steps to cancel out the common species.
First, you can use the equation C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) to find the delta H for 1 mole of CH4 (g). Since the given delta H is -376.8, you can multiply it by 1/2 to get -188.4 for 2 CH4 (g).
Next, you can use the equation C2H6(g) -> C2H4 (g) + H2 (g) with the given delta H of +136.4. Since you need 2 CH4 (g), you can multiply the delta H by 2 to get +272.8 for 2 CH4 (g).
Finally, you can add the two values obtained in the previous steps:
-188.4 (from step 1) + 272.8 (from step 2) = +84.4
Therefore, the correct answer for the value of delta H for the given reaction is +84.4, not +338.1.
3) Your answer for question 3 is correct. The value of delta H for the reaction SO3 (g) -> SO2 (g) + 1/2 O2 (g) is +99. This is because the given reaction is the reverse of the original reaction, and the value of delta H is the same magnitude but opposite in sign. Therefore, the correct answer is option a), +99.
Overall, you got two out of three answers correct.