Question

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [NO 2 ] drops from 0.0100 to 0.00650 M in 100 s. The rate of disappearance of NO2 for this period is __________ M/s.

Answer 1

Answer: The rate of disappearance of $NO_2$ is $3.5\times 10^{-5}M/s$

Explanation:

The given chemical reaction is:

$2NO_2\rightarrow 2NO+O_2$

The rate of the reaction for disappearance of $NO_2$ is given as:

$\text{Rate of disappearance of }NO_2=-\frac{\Delta [NO_2]}{\Delta t}$

Or,

$\text{Rate of disappearance of }NO_2=-\frac{C_2-C_1}{t_2-t_1}$

where,

$C_2$ = final concentration of $NO_2$ = 0.00650 M

$C_1$ = initial concentration of $NO_2$ = 0.0100 M

$t_2$ = final time = 100 minutes

$t_1$ = initial time = 0 minutes

Putting values in above equation, we get:

$\text{Rate of disappearance of }NO_2=-\frac{0.00650-0.0100}{100-0}\\\\\text{Rate of disappearance of }NO_2=3.5\times 10^{-5}M/s$

Hence, the rate of disappearance of $NO_2$ is $3.5\times 10^{-5}M/s$

Answer 2

Answer:

• The rate of  appearance of $O_2$ for this period = $1.75 * 10^{-5} M / s$

Explanation:

$rate = - \frac{\frac{1}{2} \delta [NO2]}{\delta t}\\\\rate = \frac{\frac{1}{2} \delta [NO]}{\delta t} \\\\rate = \frac{\delta [O2]}{\delta t}$

From,

$[ NO_2]_0$ = 0.01M, t1 = 0s

$[NO_2]$ = 0.00650M, t2 = 100s

Therefore,

the rate of disappearance of $NO_2$= $- \frac{\frac{1}{2} ( 0.00650 - 0.01 )}{100}$

$= 1.75 * 10^{-5} M / s$

But from the above relation rate of disappearance of $NO_2$ =  rate of appearance of $O_2$.

Therefore,

rate of appearance of $O_2$ = $1.75 * 10^{-5} M / s$

For more information on rate of disappearance, visit

https://brainly.com/subject/chemistry