Question

$N_2(g)$ reacts with $H_2(g)$ in either of the following ways depending upon supply of $H_2(g)$:

$N_2(g) + H_2(g) \longrightarrow N_2H_2(l)$

$N_2(g) + 2H_2(g) \longrightarrow N_2H_4(g)$

If 5L $N_2(g)$ and 3L $H_2(g)$ are taken initially (at same temperature and pressure), calculate the contraction in volume after the reaction (in L).

Answer 1

3

Answer 2

The problem describes two possible reactions between $N_2(g)$ and $H_2(g)$, and states that the reaction pathway depends on the supply of $H_2(g)$. We are given initial volumes of $N_2(g)$ and $H_2(g)$ at the same temperature and pressure, which means volume ratios are equivalent to mole ratios (Avogadro's Law).

Initial volumes: $V_{N_2} = 5 \, L$ $V_{H_2} = 3 \, L$

Let's analyze the two reactions:

Reaction 1: Formation of Diazene $N_2(g) + H_2(g) \longrightarrow N_2H_2(l)$

• Stoichiometry (volume basis): 1 L $N_2$ reacts with 1 L $H_2$.
• Product state: $N_2H_2$ is liquid, so its volume contribution to the gaseous phase after reaction is negligible.
• Limiting Reactant:
  • If all 5 L of $N_2$ were to react, it would require 5 L of $H_2$.
  • We only have 3 L of $H_2$. Therefore, $H_2$ is the limiting reactant.
• Reaction progress:
  • Volume of $H_2$ reacted = 3 L
  • Volume of $N_2$ reacted = 3 L (based on 1:1 ratio)
  • Volume of $N_2$ remaining = 5 L - 3 L = 2 L
• Volumes:
  • Initial total gaseous volume = $V_{N_2} + V_{H_2} = 5 \, L + 3 \, L = 8 \, L$
  • Final total gaseous volume = Volume of unreacted $N_2$ = 2 L
• Contraction in volume: Initial volume - Final volume = 8 L - 2 L = 6 L

Reaction 2: Formation of Hydrazine $N_2(g) + 2H_2(g) \longrightarrow N_2H_4(g)$

• Stoichiometry (volume basis): 1 L $N_2$ reacts with 2 L $H_2$ to produce 1 L $N_2H_4$.
• Product state: $N_2H_4$ is gaseous.
• Limiting Reactant:
  • If all 5 L of $N_2$ were to react, it would require $5 \times 2 = 10 \, L$ of $H_2$.
  • We only have 3 L of $H_2$. Therefore, $H_2$ is the limiting reactant.
• Reaction progress:
  • Volume of $H_2$ reacted = 3 L
  • Volume of $N_2$ reacted = $3 \, L / 2 = 1.5 \, L$ (based on 1:2 ratio)
  • Volume of $N_2H_4$ formed = $1.5 \, L$ (based on 1:1 ratio of $N_2$ reacted to $N_2H_4$ formed)
  • Volume of $N_2$ remaining = 5 L - 1.5 L = 3.5 L
• Volumes:
  • Initial total gaseous volume = $V_{N_2} + V_{H_2} = 5 \, L + 3 \, L = 8 \, L$
  • Final total gaseous volume = Volume of unreacted $N_2$ + Volume of $N_2H_4$ formed = 3.5 L + 1.5 L = 5 L
• Contraction in volume: Initial volume - Final volume = 8 L - 5 L = 3 L

Determining which reaction occurs: The problem states "depending upon supply of H2(g)". The initial ratio of $N_2:H_2$ is $5:3$, which is approximately $1.67:1$.

• For Reaction 1, the stoichiometric ratio $N_2:H_2$ is $1:1$. Our initial ratio of $1.67:1$ means $N_2$ is in excess, or $H_2$ is limited.
• For Reaction 2, the stoichiometric ratio $N_2:H_2$ is $1:2$. Our initial ratio of $1.67:1$ means $N_2$ is in even greater excess relative to the required $H_2$, or $H_2$ is even more limited for this reaction.

Given that the supply of $H_2$ is limited (less than what is required to react all $N_2$ in either case), it is more plausible that the reaction requiring less $H_2$ per unit of $N_2$ would be favored, or the reaction that is more likely to proceed under limited $H_2$ conditions. Reaction 1 requires 1 volume of $H_2$ per volume of $N_2$, while Reaction 2 requires 2 volumes of $H_2$ per volume of $N_2$. Therefore, Reaction 1 is favored when $H_2$ supply is limited.

Moreover, $N_2H_2$ (diazene) is known to be unstable and often disproportionates or decomposes. $N_2H_4$ (hydrazine) is a stable compound. In the absence of specific conditions, chemical reactions tend to form more stable products. This suggests Reaction 2 might be favored thermodynamically.

However, the wording "depending upon supply of H2(g)" strongly implies a consideration of stoichiometry and limiting reactants. If the supply of $H_2$ is such that it is insufficient for the higher $H_2$-demanding reaction (Reaction 2) to consume all $N_2$, then the reaction that is more "feasible" with the available $H_2$ or results in a more significant volume change might be expected.

Let's re-evaluate. If the problem implies that the reaction pathway is chosen to consume the available $H_2$ as efficiently as possible towards a stable product, Reaction 2 might be considered.

Let's consider the "contraction in volume" aspect. If Reaction 1 occurs, contraction = 6 L. If Reaction 2 occurs, contraction = 3 L.

Usually, in such competitive reactions, the one that is thermodynamically more favorable (forming a more stable product) or kinetically faster is chosen. Hydrazine ($N_2H_4$) is a stable compound, while diazene ($N_2H_2$) is unstable. This would favor Reaction 2. Also, the phrase "depending upon supply of H2(g)" might imply that if enough H2 is present to form the more reduced product (hydrazine, $N_2H_4$), then that reaction occurs. Since 3L of H2 can react to form $N_2H_4$ (it just won't consume all the N2), it's a possibility.

Given the common context of such problems in competitive exams, if there's a choice between forming a stable gaseous product vs. an unstable liquid/solid product, and both are stoichiometrically possible to some extent, the stable gaseous product is often the intended answer. Let's assume Reaction 2 is the intended reaction.

Final Calculation for Reaction 2: Initial total gaseous volume = 8 L Final total gaseous volume = 5 L Contraction in volume = 8 L - 5 L = 3 L