Question

Nitric oxide reacts with \( \text{H}_2 \) according to the reaction: 
\[ 2\text{NO}_{\text{(g)}} + 2\text{H}_2{}_{\text{(g)}} \rightarrow \text{N}_2{}_{\text{(g)}} + 2\text{H}_2\text{O}_{\text{(g)}} \] Identify the correct relation for the disappearance of reactants and appearance of products.

• A. \(\frac{1}{2} \frac{\text{d}[\text{NO}]}{\text{dt}} = -\frac{\text{d}[\text{H}_2]}{\text{dt}}\)
• B. \(\frac{\text{d}[\text{N}_2]}{\text{dt}} = \frac{1}{2} \frac{\text{d}[\text{H}_2\text{O}]}{\text{dt}}\)
• C. \(-\frac{\text{d}[\text{N}_2]}{\text{dt}} = \frac{1}{2} \frac{\text{d}[\text{H}_2\text{O}]}{\text{dt}}\)
• D. \(\frac{\text{d}[\text{H}_2]}{\text{dt}} = -\frac{1}{2} \frac{\text{d}[\text{N}_2]}{\text{dt}}\)

Hint:

For rate relations, always divide by stoichiometric coefficients and use negative sign for reactants, positive sign for products.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The rate of a chemical reaction can be expressed in terms of the rate of disappearance of any reactant or the rate of appearance of any product.
To equate these specific rates to a single overall "Rate of Reaction", we divide each specific rate by its corresponding stoichiometric coefficient from the balanced chemical equation.
Reactants are assigned a negative sign because their concentration decreases, while products are assigned a positive sign.
Step 2: Key Formula or Approach:
The general rate expression for \(aA + bB \rightarrow cC + dD\) is:
\[ \text{Rate} = -\frac{1}{a}\frac{\text{d}[A]}{\text{dt}} = -\frac{1}{b}\frac{\text{d}[B]}{\text{dt}} = +\frac{1}{c}\frac{\text{d}[C]}{\text{dt}} = +\frac{1}{d}\frac{\text{d}[D]}{\text{dt}} \] Step 3: Detailed Explanation:
Given the balanced chemical equation:
\[ 2\text{NO}_{\text{(g)}} + 2\text{H}_{2\text{(g)}} \rightarrow \text{N}_{2\text{(g)}} + 2\text{H}_2\text{O}_{\text{(g)}} \] Using the rule above, we can write the overall rate expression as:
\[ \text{Rate} = -\frac{1}{2}\frac{\text{d}[\text{NO}]}{\text{dt}} = -\frac{1}{2}\frac{\text{d}[\text{H}_2]}{\text{dt}} = +\frac{1}{1}\frac{\text{d}[\text{N}_2]}{\text{dt}} = +\frac{1}{2}\frac{\text{d}[\text{H}_2\text{O}]}{\text{dt}} \] Now, let's analyze the given options to find the correct equality:
Option (A): \(\frac{1}{2} \frac{\text{d}[\text{NO}]}{\text{dt}} = -\frac{\text{d}[\text{H}_2]}{\text{dt}}\). From our rate expression, \(-\frac{1}{2}\frac{\text{d}[\text{NO}]}{\text{dt}} = -\frac{1}{2}\frac{\text{d}[\text{H}_2]}{\text{dt}}\), which simplifies to \(\frac{\text{d}[\text{NO}]}{\text{dt}} = \frac{\text{d}[\text{H}_2]}{\text{dt}}\). Thus, (A) is incorrect.
Option (B): \(\frac{\text{d}[\text{N}_2]}{\text{dt}} = \frac{1}{2} \frac{\text{d}[\text{H}_2\text{O}]}{\text{dt}}\). This exactly matches the equality \(+\frac{1}{1}\frac{\text{d}[\text{N}_2]}{\text{dt}} = +\frac{1}{2}\frac{\text{d}[\text{H}_2\text{O}]}{\text{dt}}\) from our derived rate expression. Thus, (B) is correct.
Option (C): \(-\frac{\text{d}[\text{N}_2]}{\text{dt}} = \frac{1}{2} \frac{\text{d}[\text{H}_2\text{O}]}{\text{dt}}\). Nitrogen (\(\text{N}_2\)) is a product, so its appearance rate must be positive when equated to the appearance rate of another product. The negative sign makes it incorrect.
Option (D): \(\frac{\text{d}[\text{H}_2]}{\text{dt}} = -\frac{1}{2} \frac{\text{d}[\text{N}_2]}{\text{dt}}\). From our expression, \(-\frac{1}{2}\frac{\text{d}[\text{H}_2]}{\text{dt}} = \frac{\text{d}[\text{N}_2]}{\text{dt}}\), which means \(\frac{\text{d}[\text{H}_2]}{\text{dt}} = -2\frac{\text{d}[\text{N}_2]}{\text{dt}}\). Thus, (D) is incorrect.
Step 4: Final Answer:
The correct relation is given in option (B).