2NO2(g)= 2NO(g) + O2(g) Nitrogen dioxide is dissociated to the extent of 56.6 percent and 494°C and 99kPa pressure. At what pressure will the dissociation be 80 percent at 494°C?

AT THE END OF 80% DISSOLUTION, THE PRESSURE OF NO2 HAS CHANGED FROM 99kPa TO 139.97kPa

Explanation:

P1 = 99 kPa

P2 = unknown

From the reaction,

2 mole of NO2 will produce 2 mole of NO

We can also say that 1 mole of NO2 will produce 1 mole of NO

At 56.6 % of NO2, 0.566 mole of NO2 will be consumed

At STP, 1 mole of  a substance will occupy 22.4 dm3 volume

0.566 mole will occupy ( 22.4 * 0.566 / 1) dm3 volume

= 39.58 dm3 volume

V1 = 39.56 dm3

At the new percent of 80%, 0.80 mole of NO2 will be consumed

Since, 1 mole = 22.4 dm3

0.80 mole = (22.4 / 0.80) dm3

= 28 dm3

V2 = 28 dm3

Using the equation of Boyle's law which shows the relationship between pressure and volume of a given mass of gas at constant temperature, we have:

P1 V1 = P2 V2

Re-arranging to make P2 the subject of formula:

P2 = P1V1 / V2

P2 = 99 kPa * 39.56 / 28

P2 = 3916.44 kPa / 28

P2 = 139.87 kPa

So at 80 % dissociation of NO2, the pressure has changed from 99 kPa to 139.97 kPa.

you didnt put any of the following

c) should be the answer.