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Drupady [299]
3 years ago
11

Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and

volume were constant at values of 303 K and 2.00 L, respectively. If the pressure was 10.4 atm prior to the reaction, what would be the expected pressure after the reaction was completed?
Chemistry
1 answer:
andrew11 [14]3 years ago
7 0

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454
}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

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