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What effect does a decrease in the concentration of nitrogen monoxide [tex]\((NO)\)[/tex] gas have on the rate of reaction in this chemical equation?

[tex]\[ 2NO(g) + H_2(g) \rightarrow N_2O(g) + H_2O(g) \][/tex]

A decrease in the concentration of nitrogen monoxide [tex]\(\square\)[/tex] between [tex]\(NO\)[/tex] and [tex]\(H_2\)[/tex] molecules. The rate of the forward reaction then [tex]\(\square\)[/tex].

Sagot :

GinnyAnsweridn
Let's analyze the provided chemical reaction to understand the impact of a change in concentration of one of the reactants. The reaction is:

[tex]\[ 2 \text{NO} (g) + \text{H}_2 (g) \rightarrow \text{N}_2\text{O} (g) + \text{H}_2\text{O} (g) \][/tex]

We need to determine what happens when the concentration of nitrogen monoxide (NO) decreases.

1. Effect on Collision Frequency:
- According to the law of mass action, the rate of a chemical reaction is proportional to the product of the concentrations of the reactants. In this case, the reaction involves nitrogen monoxide (NO) and hydrogen gas (H[tex]$_2$[/tex]).
- If the concentration of NO decreases, there will be fewer NO molecules available in the system.
- Since there are fewer NO molecules, the frequency of collisions between NO molecules and H[tex]$_2$[/tex] molecules will decrease.

2. Effect on Reaction Rate:
- The rate of the forward reaction depends on how frequently the reactant molecules collide with enough energy to react. A decrease in the concentration of NO means that there are fewer successful collisions occurring between NO and H[tex]$_2$[/tex] molecules.
- Therefore, the overall rate of the forward reaction will decrease as well.

Based on this detailed explanation, we can fill in the blanks for the question:

- A decrease in the concentration of nitrogen monoxide decreases collisions between NO and H[tex]$_2$[/tex] molecules.
- The rate of the forward reaction then decreases.

So, the correct answer will be:
```
A decrease in the concentration of nitrogen monoxide decreases collisions between NO and H _2 molecules. The rate of the forward reaction then decreases.
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