Question

In this reaction, a gaseous system with a volume of 3.00 L has a rate of formation of NOCl of 0.0120 M s-1 . The volume of the system is decreased to 1.00 L with no change of temperature. What is the new rate of formation of NOCl?

Answer 1

x = 0.324 M s⁻¹

Step-by-step explanation:

Equation for the reaction can be represented as:

2 NO(g) + Cl₂ (g) ⇄ 2NOCl (g)

Rate = K [NO]² [Cl₂]

Concentration =
`(numbers of mole (n))/(volume (v))`

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 ∝
`(1)/(v)`

Concentration =
`(1)/(v)`

Similarly; the Rate can now be expressed as:

Rate = K [NO]² [Cl₂]

Rate =
`((1)/(v)) ^2`
`((1)/(v) )`

Rate =
`(1)/(v^3)`

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 =
`(1)/(3^3)`

0.0120 =
`(1)/(27)` -----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 =
`((1)/(1))^3`

x = 1 ------- Equation (2)

Dividing equation (2) with equation (1); we have:

`(0.0210)/(x)` =
`((1)/(27) )/(1)`

`(0.0210)/(x)` =
`(1)/(27)`

x = 0.0120 × 27

x = 0.324 M s⁻¹

∴ the new rate of formation of NOCl = 0.324 M s⁻¹