Question

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?

Answer 1

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.