Answer:
x = 0.324 M s⁻¹
Explanation:
Equation for the reaction can be represented as:
2 NO(g) + Cl₂ (g) ⇄ 2NOCl (g)
Rate = K [NO]² [Cl₂]
Concentration = [tex]\frac{numbers of mole (n)}{volume (v)}[/tex]
from the question; their number of moles are constant since the species are quite alike.
As such; if Concentration varies inversely proportional to the volume;
we have: Concentration ∝ [tex]\frac{1}{v}[/tex]
Concentration = [tex]\frac{1}{v}[/tex]
Similarly; the Rate can now be expressed as:
Rate = K [NO]² [Cl₂]
Rate = [tex](\frac{1}{v}) ^2[/tex] [tex](\frac{1}{v} )[/tex]
Rate = [tex]\frac{1}{v^3}[/tex]
We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹
So we can have:
0.0120 = [tex]\frac{1}{3^3}[/tex]
0.0120 = [tex]\frac{1}{27}[/tex] -----Equation (1)
Now; the new rate of formation when the volume of the system decreased to 1.00 L can now be calculated as:
x = [tex](\frac{1}{1})^3[/tex]
x = 1 ------- Equation (2)
Dividing equation (2) with equation (1); we have:
[tex]\frac{0.0210}{x}[/tex] = [tex]\frac{\frac{1}{27} }{1}[/tex]
[tex]\frac{0.0210}{x}[/tex] = [tex]\frac{1}{27}[/tex]
x = 0.0120 × 27
x = 0.324 M s⁻¹
∴ the new rate of formation of NOCl = 0.324 M s⁻¹
