- #1

Heresy42

- 5

- 0

Dear all,

For practice, I'm trying to solve an exercise I've found on the Internet. A screen shot of it is attached. I would like to be sure about my approach, especially when it comes to question 2.

Let's say, for the sake of simplification, that the reaction is as follows (no consideration of dissociation of ammonia):

5 N2H4 (g) -> 4 NH2 + 3 N2 + 6H2

1/ Heat release per kg of hydrazine? I was thinking of computing the enthalpy of formation of the reaction to begin with: Delta Hf = 4 Delta Hf NH2 + 3 Delta Hf N2 + 6 Delta Hf H2 - 5 Delta Hf N2H4. The next step is to multiply by 5 mole and divide by 1 kg, right?

2/ Temperature of the products of decomposition, called T? What about the following, assuming that all hydrazine has been consumed: [tex] \Delta \ Hf (\mathrm{previously \ computed}) = 4 \int_{25}^{T} cp_{NH2} dT + 3 \int_{25}^{T} cp_{N2} dT + 6 \int_{25}^{T} cp_{H2} dT - 5 \int_{25}^{25} cp_{NH2} dT. [/tex]

Is this equation right? Because I don't really how to tackle this question. What about the sign? Is it "Delta Hf ="or "- Delta Hf =".

3/ Molecular mass of the products MW? Is it possible to use the simple formula: MW = MWNH2*(4/13) + MWN2*(3/13) + MWH2*(6/13), with 13 = 4 + 3 + 2, assuming that all hydrazine has been consumed?

Thank you for checking my reasoning.

Regards.

## Homework Statement

For practice, I'm trying to solve an exercise I've found on the Internet. A screen shot of it is attached. I would like to be sure about my approach, especially when it comes to question 2.

Let's say, for the sake of simplification, that the reaction is as follows (no consideration of dissociation of ammonia):

5 N2H4 (g) -> 4 NH2 + 3 N2 + 6H2

**2. The attempt at a solution**1/ Heat release per kg of hydrazine? I was thinking of computing the enthalpy of formation of the reaction to begin with: Delta Hf = 4 Delta Hf NH2 + 3 Delta Hf N2 + 6 Delta Hf H2 - 5 Delta Hf N2H4. The next step is to multiply by 5 mole and divide by 1 kg, right?

2/ Temperature of the products of decomposition, called T? What about the following, assuming that all hydrazine has been consumed: [tex] \Delta \ Hf (\mathrm{previously \ computed}) = 4 \int_{25}^{T} cp_{NH2} dT + 3 \int_{25}^{T} cp_{N2} dT + 6 \int_{25}^{T} cp_{H2} dT - 5 \int_{25}^{25} cp_{NH2} dT. [/tex]

Is this equation right? Because I don't really how to tackle this question. What about the sign? Is it "Delta Hf ="or "- Delta Hf =".

3/ Molecular mass of the products MW? Is it possible to use the simple formula: MW = MWNH2*(4/13) + MWN2*(3/13) + MWH2*(6/13), with 13 = 4 + 3 + 2, assuming that all hydrazine has been consumed?

Thank you for checking my reasoning.

Regards.