Question

In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

Answer 1

The question is incomplete, here is the complete question:

Consider the following reaction:  $2N_2O(g)\rightarrow 2N_2(g)+O_2(g)$

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

Answer: The average rate of appearance of oxygen gas is $2.64\times 10^{-3}M/s$

Explanation:

We are given:

Moles of oxygen gas = $1.9\times 10^{-2}moles$

Volume of solution = 0.480 L

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

So, $\text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M$

The given chemical reaction follows:

$2N_2O(g)\rightarrow 2N_2(g)+O_2(g)$

The average rate of the reaction for appearance of $O_2$ is given as:

$\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}$

Or,

$\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}$

where,

$C_2$ = final concentration of oxygen gas = 0.0396 M

$C_1$ = initial concentration of oxygen gas = 0 M

$t_2$ = final time = 15.0 s

$t_1$ = initial time = 0 s

Putting values in above equation, we get:

$\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s$

Hence, the average rate of appearance of oxygen gas is $2.64\times 10^{-3}M/s$