To understand how the reaction rate changes when the concentrations of H₂ and NO are both doubled, let's break down the problem step by step.
### Given Information:
- The reaction is: [tex]\(2 \text{NO} (g) + 2 \text{H}_2 (g) \rightarrow \text{N}_2 (g) + 2 \text{H}_2\text{O} (g)\)[/tex]
- Rate law orders:
- First order in [tex]\( \text{H}_2 \)[/tex]
- Second order in [tex]\( \text{NO} \)[/tex]
### Rate Law Expression:
The rate law for this reaction can be expressed as:
[tex]\[ \text{Rate} = k [\text{H}_2]^1 [\text{NO}]^2 \][/tex]
where [tex]\( k \)[/tex] is the rate constant.
### Initial Conditions:
Let the initial concentrations of [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{NO} \)[/tex] be [tex]\( [\text{H}_2] \)[/tex] and [tex]\( [\text{NO}] \)[/tex]. The initial rate [tex]\( \text{Rate}_{\text{initial}} \)[/tex] can be written as:
[tex]\[ \text{Rate}_{\text{initial}} = k [\text{H}_2] [\text{NO}]^2 \][/tex]
### New Conditions with Doubled Concentrations:
When the concentrations of [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{NO} \)[/tex] are both doubled, the new concentrations become [tex]\( 2[\text{H}_2] \)[/tex] and [tex]\( 2[\text{NO}] \)[/tex].
The new rate [tex]\( \text{Rate}_{\text{new}} \)[/tex] with the doubled concentrations is:
[tex]\[ \text{Rate}_{\text{new}} = k (2[\text{H}_2])(2[\text{NO}])^2 \][/tex]
Expanding this, we get:
[tex]\[ \text{Rate}_{\text{new}} = k (2[\text{H}_2])(4[\text{NO}]^2) \][/tex]
[tex]\[ \text{Rate}_{\text{new}} = k (2) [\text{H}_2] (4) [\text{NO}]^2 \][/tex]
[tex]\[ \text{Rate}_{\text{new}} = 8 \cdot k [\text{H}_2] [\text{NO}]^2 \][/tex]
[tex]\[ \text{Rate}_{\text{new}} = 8 \cdot \text{Rate}_{\text{initial}} \][/tex]
### Conclusion:
When the concentrations of both [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{NO} \)[/tex] are doubled, the rate of the reaction increases by a factor of 8.
Thus, the correct answer is:
- The rate increases by a factor of 8.
