Question

For reaction \( N _{2}( g )+3 H _{2}( g ) \rightarrow 2 NH _{3}( g ), 1 mol N_2 \) and \( 4 mol H_2 \) are taken in \( 15 L \) flask at \( 27^{\circ} C \). After complete conversion of \( N _{2} \) into \( NH _{3}, 5 L \) of \( H _{2} O \) is added. Pressure set up in the flask is 

a) \( \frac{3 \times 0.0821 \times 300}{15} atm \) 

b) \( \frac{2 \times 0.0821 \times 300}{10} atm \) 

c) \( \frac{1 \times 0.0821 \times 300}{15} atm \) 

d) \( \frac{3 \times 0.0821 \times 300}{10} atm \)

Answer 1

Correct option is (d) \(\cfrac{3\times0.0821\times300}{10}atm\)

We have, numbers of moles of N₂ = 1 mol

Numbers of moles H₂ = 4 mol

From the above chemical equation, we can see that  1 mol of N₂ react with 3 mol of Hydrogen to form 2 mol NH₃ gas

We have, one mole N₂ and 4 mol H₂. It means 1 mole H₂ Hydrogen gas remains unreacted after complete reaction.

\(\therefore\) Total numbers of moles of gases = 1 mol H₂ + 2 mol NH₃ = 3 moles

After addition of 5 L of water the volumes of flask reduced to 10 L.

Using ideal gas equation

PV = nRT (Assuming H₂ and NH₃ are not dissolve in water)

P = \(\cfrac{3\times0.0821\times300}{10}atm\)