a certain reaction is second order in n₂ and first order in h₂. use this information to complete the table…
a certain reaction is second order in n₂ and first order in h₂. use this information to complete the table below. round each of your answers to 3 significant digits.\n| n₂ | h₂ | initial rate of reaction |\n| ---- | ---- | ---- |\n| 2.26 m | 1.08 m | 55.0 m/s |\n| 2.26 m | 0.396 m | m/s |\n| 0.260 m | 9.40 m | m/s |
Answer
Explanation:
Step1: Determine the rate - law expression
The rate - law for the reaction is $r = k[\text{N}_2]^2[\text{H}_2]$. First, find the rate constant $k$ using the first set of data. Given $[\text{N}_2]=2.26\ M$, $[\text{H}_2]=1.08\ M$ and $r = 55.0\ M/s$. [k=\frac{r}{[\text{N}_2]^2[\text{H}_2]}=\frac{55.0}{(2.26)^2\times1.08}] [k=\frac{55.0}{2.26^2\times1.08}=\frac{55.0}{5.55}\approx9.91\ M^{-2}s^{-1}]
Step2: Calculate the rate for the second row
For the second row, $[\text{N}_2]=2.26\ M$ and $[\text{H}_2]=0.396\ M$. Using the rate - law $r = k[\text{N}_2]^2[\text{H}_2]$ and $k = 9.91\ M^{-2}s^{-1}$. [r=9.91\times(2.26)^2\times0.396] [r=9.91\times5.1076\times0.396] [r=9.91\times2.0226\approx20.0\ M/s]
Step3: Calculate the rate for the third row
For the third row, $[\text{N}_2]=0.260\ M$ and $[\text{H}_2]=9.40\ M$. Using the rate - law $r = k[\text{N}_2]^2[\text{H}_2]$ and $k = 9.91\ M^{-2}s^{-1}$. [r=9.91\times(0.260)^2\times9.40] [r=9.91\times0.0676\times9.40] [r=9.91\times0.63544\approx6.30\ M/s]
Answer:
| $[\text{N}_2]$ | $[\text{H}_2]$ | initial rate of reaction |
|---|---|---|
| $2.26\ M$ | $1.08\ M$ | $55.0\ M/s$ |
| $2.26\ M$ | $0.396\ M$ | $20.0\ M/s$ |
| $0.260\ M$ | $9.40\ M$ | $6.30\ M/s$ |