Question

Given the enthalpies of reaction below, determine the enthalpy change for the reaction: N2O (g) + NO2 (g) 3NO (g).
N2 (g) + O2 (g) 2NO (g)
ΔH = +180.7 kJ

2NO (g) + O2 (g) 2NO2 (g)
ΔH = -113.1 kJ

2N2O (g) 2N2 (g) + O2 (g)
ΔH = -163.2 kJ


A. 155.6 kJ
B. 130.6 kJ
C. 311.3 kJ
D. 65.3 kJ

Answer 1

Do you know the summation equation for DH? Something like:\[\Delta H_{rxn} = \Sigma \Delta H_{products} - \Sigma \Delta H_{reactants}\]

Answer 2

The goal is to use the above equations and make it so that you only have the first reaction that you stated remaining.

Answer 3

You have:

N2O + NO2 --> 3 NO

^- you want to manipulate the below equations so you end up getting the above equation:

[1] N2 + O2 --> 2NO

[2] 2NO + O2 --> 2NO2

[3] 2N2O --> 2N2 + 1O2

Hint:
- you should try getting rid of the following:
a) O2 (try reversing one of the equations)
b) N2 (try reversing and multiply by 2, OR dividing by 2 one of the equations)

- You know that you have "N2O" in the final equation, so in [3], what should you do with "2N2O"?

Answer 4

- The goal is to manipulate the three equations so that you end up getting the top equation (one i wrote without a number)

NOTE:
- Suppose you reverse an equation (for instance: [3]). Then, you must muliply the corresponding ΔH by -1.

- If you multiply the equation by anything (for example: 3), then you must multiply the corresponding ΔH by that number.

Answer 5

So, could you explain that in a way where someone who doesn't know anything about chemistry would understand?

Answer 6

[1] N2 + O2 --> 2NO

[2] 2NO + O2 --> 2NO2

[3] 2N2O --> 2N2 + 1O2


We want to get: [*] N2O + NO2 --> 3 NO

Suppose we have a generic equation:
[&] A + B --> C, with ΔH = 20kJ

1. REVERSAL
We may "reverse" this equation as follows:
[&] C --> A + B

But then, we also have to "reverse" the ΔH. This is done by multiplying it by -1.

So, we end up getting:
[&] C --> A + B, with ΔH = -20kJ

-----------------------------

2. MULTIPLICATION
We may multiple the equation by any number. Suppose we multiply the equation by 2:
[&] A + B --> C, with ΔH = 20kJ

becomes:

[&] 2(A + B) --> 2(C)
- remember, what we do to one side, we also have to do to the other side

so it becomes:
[&] 2A + 2B --> 2C

But then, we also have to multiply the ΔH by the same number. So, we will ultimately get:

[&] 2A + 2B --> 2C, ΔH = 40kJ

------------------------------

CANCELLATION AND ADDITION

Suppose we have:
[#1] A + B --> C + E, with ΔH = 20kJ
[#2] C + D --> A, with ΔH = 20kJ

We can add these equations like so:

A + B + C + D --> C + A + E
Then, we can "cancel" the letters that are common to both sides. So in this case, it is C and A. Then, we end up getting:

B + D --> E

But, again we also have to add the ΔH (ie. the enthalpies) as well:
ΔH = 20 + 20 = 40kJ

Moral of the story:
- what you do with an equation, you must also do so with the enthalpy (or ΔH)

Answer 7

Thank you

Answer 8

No problem.

Answer 9

Ask me, or anyone else available if you ever need help again.

Answer 10

so what is the answer?