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Snezhnost [94]
3 years ago
11

1. Calculate AH for the reaction

Chemistry
1 answer:
Katarina [22]3 years ago
5 0

Answer:

- 58 kJ

Explanation:

The equation of the reaction is given as;

2NO2(g) → N2O4(g)

N2(g) + 202(g) → 2NO2(g) delta H = 67.7 kJ

Since NO2 is the reactant in the equation, we have to reverse this reaction. We now have;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g) delta H = 9.7 kJ

Since NO4 is the product in the equation, we don't have to do anything to this reaction.

Adding the two reactions;

2NO2(g) → N2(g) + 202(g)     delta H = - 67.7 kJ

N2(g) + 202(g) → N2O4(g)     delta H = 9.7 kJ

---------------------------------------------------------------- N2(g) and 2O2 (g) cancels out

2NO2(g) → N2O4(g)       delta H = (-67.7 + 9.7) = - 58 kJ

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See below ~

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Explanation:

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10. If 200. L of ammonia (NH3) are produced in the reaction between nitrogen and hydrogen at STP, how many
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Answer:

100L of N2

Explanation:

First let us obtain the number of mole of NH3 that occupied 200L at stp. This can be achieved by doing the following:

1mole of NH3 occupied 22.4L at stp.

Therefore, Xmol mole of NH3 will occupy 200L i.e

Xmol of NH3 = 200/22.4 = 8.93moles.

Now let us generate a balanced equation for the reaction.

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From the equation above,

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Therefore, Xmol of N2 will produce 8.93moles of NH3 i.e

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Therefore, 4.47moles of N2 will occupy = 4.47 x 22.4 = 100L

Therefore, 100L of N2 is used in the reaction

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