Question

For the reaction $\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}$, what is the relation between $\frac{d[\text{N}_2]}{dt}$ and $\frac{d[\text{H}_2]}{dt}$?

• A. $3 \frac{d[\text{H}_2]}{dt} = \frac{d[\text{N}_2]}{dt}$
• B. $\frac{d[\text{H}_2]}{dt} = 3 \frac{d[\text{N}_2]}{dt}$
• C. $3 \frac{d[\text{H}_2]}{dt} = 2 \frac{d[\text{N}_2]}{dt}$
• D. $\frac{d[\text{H}_2]}{dt} = \frac{d[\text{N}_2]}{dt}$

Hint:

You can set up these rate equations instantly without writing the full fractions! Just think of it intuitively: because the stoichiometry for $\text{H}_2$ is 3 times larger than $\text{N}_2$, hydrogen must disappear exactly 3 times faster than nitrogen. Therefore: $\text{Rate of }\text{H}_2 = 3 \times (\text{Rate of }\text{N}_2)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question requires us to establish a relative mathematical correlation between the rates of consumption of two different reactants ($\text{N}_2$ and $\text{H}_2$) in the synthesis of ammonia.

Step 2: Key Formula or Approach:
For a general homogeneous gaseous reaction of the type $aA + bB \rightarrow cC$, the overall instantaneous rate of reaction is written by dividing each rate of change by its corresponding stoichiometric coefficient: $$\text{Rate of reaction} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt}$$ The negative sign indicates that the reactants are consumed over time.

Step 3: Detailed Explanation:
Let's write down the full rate expression for our specific chemical equilibrium: $$\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}$$ $$\text{Rate of reaction} = -\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt} = +\frac{1}{2}\frac{d[\text{NH}_3]}{dt}$$ To isolate the relationship between the consumption of nitrogen and hydrogen, we equate their individual terms: $$-\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt}$$ The negative signs cancel out: $$\frac{d[\text{N}_2]}{dt} = \frac{1}{3}\frac{d[\text{H}_2]}{dt}$$ Rearranging this to match the formatting of the options gives: $$\frac{d[\text{H}_2]}{dt} = 3 \frac{d[\text{N}_2]}{dt}$$

Step 4: Final Answer:
The correct relation is $\frac{d[\text{H}_2]}{dt} = 3 \frac{d[\text{N}_2]}{dt}$, which matches option (B).