Question

consider this balanced equation:

2n₂o(g)→2n₂(g)+o₂(g)

express the rate of the reaction in terms of the change in concentration of each of the reactants and products

in the first 15.0 seconds of the reaction, 0.015 mol of o₂ is produced in a reaction vessel with a volume of 0.500 l. what is the average rate of reaction during this time interval?

predict the rate of change of n₂o during this time interval. in other words, what is δn₂o/δt?

Explanation:

Step1: Express reaction rate

For a general reaction \(aA + bB
ightarrow cC + dD\), the rate is \(-\frac{1}{a}\frac{\Delta[A]}{\Delta t}=-\frac{1}{b}\frac{\Delta[B]}{\Delta t}=\frac{1}{c}\frac{\Delta[C]}{\Delta t}=\frac{1}{d}\frac{\Delta[D]}{\Delta t}\).
For the reaction \(2N_{2}O(g)
ightarrow 2N_{2}(g)+O_{2}(g)\), the rate \(r =-\frac{1}{2}\frac{\Delta[N_{2}O]}{\Delta t}=\frac{1}{2}\frac{\Delta[N_{2}]}{\Delta t}=\frac{\Delta[O_{2}]}{\Delta t}\).

Step2: Calculate \(\Delta[O_{2}]\)

Concentration \(c=\frac{n}{V}\). Given \(n = 0.015\space mol\) and \(V=0.500\space L\), \(\Delta[O_{2}]=\frac{0.015\space mol}{0.500\space L}=0.030\space M\).
\(\Delta t = 15.0\space s\).
The average rate of reaction \(r=\frac{\Delta[O_{2}]}{\Delta t}\).
Substitute values: \(r=\frac{0.030\space M}{15.0\space s}=2.0\times10^{-3}\space M/s\).

Step3: Find \(\frac{\Delta[N_{2}O]}{\Delta t}\)

From \(r =-\frac{1}{2}\frac{\Delta[N_{2}O]}{\Delta t}\), we can solve for \(\frac{\Delta[N_{2}O]}{\Delta t}\).
\(\frac{\Delta[N_{2}O]}{\Delta t}=- 2r\).
Substitute \(r = 2.0\times10^{-3}\space M/s\): \(\frac{\Delta[N_{2}O]}{\Delta t}=-4.0\times10^{-3}\space M/s\).

Answer:

• Rate expression: \(r =-\frac{1}{2}\frac{\Delta[N_{2}O]}{\Delta t}=\frac{1}{2}\frac{\Delta[N_{2}]}{\Delta t}=\frac{\Delta[O_{2}]}{\Delta t}\).
• Average rate of reaction: \(2.0\times 10^{-3}\space M/s\).
• \(\frac{\Delta[N_{2}O]}{\Delta t}=-4.0\times 10^{-3}\space M/s\).