Answer: The partial pressure of nitrogen dioxide after it is reached the second time is 1.454 atm
Step-by-step explanation:
We are given:
Initial partial pressure of nitrogen dioxide gas = 1.9 atm
First equilibrium partial pressure of nitrogen dioxide gas = 1.1 atm
The chemical equation for the reaction of nitrogen dioxide to dinitrogen tetroxide follows:

Initial: 1.9
At eqllm: 1.9-2x x
Evaluating the value of 'x'

So, equilibrium partial pressure of dinitrogen tetroxide = x = 0.4 atm
The expression of
for above equation follows:
..........(1)
Putting values in above expression, we get:

Now, when more nitrogen dioxide is added, equilibrium is re-established:
Partial pressure of nitrogen dioxide added = 0.95 atm

Initial: 2.05 0.4
At eqllm: 2.05-2x 0.4+x
Putting values in expression 1, we get:

Neglecting the value of x = 2.51 because the equilibrium partial pressure of nitrogen dioxide will become negative, which is not possible.
So, the equilibrium partial pressure of nitrogen dioxide after it is reached the second time = (2.05 - 2x) = [2.05 - 2(0.298)] = 1.454 atm
Hence, the partial pressure of nitrogen dioxide after it is reached the second time is 1.454 atm