Respuesta :
The question is incomplete, here is the complete question:
Consider the following reaction: [tex]2N_2O(g)\rightarrow 2N_2(g)+O_2(g)[/tex]
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 [tex]2.64\times 10^{-3}M/s[/tex]
Explanation:
We are given:
Moles of oxygen gas = [tex]1.9\times 10^{-2}moles[/tex]
Volume of solution = 0.480 L
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
So, [tex]\text{Molarity of }O_2=\frac{1.9\times 10^{-2}mol}{0.480L}=0.0396M[/tex]
The given chemical reaction follows:
[tex]2N_2O(g)\rightarrow 2N_2(g)+O_2(g)[/tex]
The average rate of the reaction for appearance of [tex]O_2[/tex] is given as:
[tex]\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}[/tex]
Or,
[tex]\text{Average rate of appearance of }O_2=\frac{C_2-C_1}{t_2-t_1}[/tex]
where,
[tex]C_2[/tex] = final concentration of oxygen gas = 0.0396 M
[tex]C_1[/tex] = initial concentration of oxygen gas = 0 M
[tex]t_2[/tex] = final time = 15.0 s
[tex]t_1[/tex] = initial time = 0 s
Putting values in above equation, we get:
[tex]\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[/tex]
Hence, the average rate of appearance of oxygen gas is [tex]2.64\times 10^{-3}M/s[/tex]
