[Nitin_Naudiyal]

Please help me with the following question

if the volume of the vessel in which the reaction 2NO + O₂  2NO₂ is occurring is diminished to 1/3rd of its initial volume. The rate of reaction will be increased by?

Solution:
As Pressure increases, the Rate of Reaction also increases.
By Ideal Gas Law,
P = n / V ( RT - constant at constant temperature)

So, if i consider that there are 3 Moles of Reactant and the volume is diminished by 1/3rd.
The increase in P should be 9 times and Hence, the Rate of Reaction increase by 9 times.
However, the answer given is Rate of Reaction will be increased by 27.

Where did i go wrong?.

Thank you.

[Borek]

Write expression for the reaction rate.

[Nitin_Naudiyal]

The Rate of Reaction = -(1/2)d[NO]/dt = - d[O₂]/dt = (1/2)d[NO₂]/dt

[Nitin_Naudiyal]

I got a solution

For the Equation, 2NO + O₂    2NO₂

The Rate Law = k [NO]²[O₂]
So, if the volume of the vessel is diminised to 1/3rd, the concentration will increase by 3 times.

Hence, New Rate = k [3NO]²[3O₂] = 27k [NO]²[O₂]

And Hence, the rate will increase by 27 times.

However, it is given that, the rate law cannot be determined from the stiochometric equation.
So, is this method the correct approach.