Answer: The concentration of nitrogen dioxide at equilibrium is 0.063 M
Step-by-step explanation:
We are given:
Initial concentration of nitrogen dioxide = 0.0250 M
Initial concentration of dinitrogen tetraoxide = 0.0250 M
For the given chemical equation:

Initial: 0.025 0.025
At eqllm: 0.025-x 0.025+2x
The expression of
for above equation follows:
![K_c=([NO_2]^2)/([N_2O_4])](https://img.qammunity.org/2021/formulas/chemistry/college/vkb4qh8uu2trux626hsyhcrfq5sgbbyvg6.png)
We are given:

Putting values in above expression, we get:

Neglecting the negative value of 'x', because concentration cannot be negative
So, equilibrium concentration of nitrogen dioxide = (0.025 + 2x) = [0.025 + 2(0.019)] = 0.063 M
Hence, the concentration of nitrogen dioxide at equilibrium is 0.063 M