Test Index

NCERT Class XI Chemistry Equilibrium Solutions

© examsnet.com
Question : 8
Total: 73
Reaction between N2 and O2 takes place as follows :
2N2(g)+O2(g) ⇌ 2N2O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
Solution:  
2N2(g)+O2(g) ⇌ 2N2O(g)
Initial‌no.of‌moles:0.4820.9330
At‌eqm.no.of‌moles:(0.482−x)(0.933−
x
2
)		x
Molar‌conc.
0.482−x
10
0.933−(
x
2
)
10
x
10

As Kc = 2.0 × 10–37 is very small, this means that the amount of N2 and O2reacted (x) is very very small. Hence, at equilibrium, we have[N2] = 0.0482 molL–1,[O2] = 0.0933 molL–1,[N2O] = 0.1x
∴ Kc =
(0.1x)2
(0.0482)2(0.0933)
 = 2.0 × 10−37 (given)
On solving this gives, x = 6.6 × 10–20
∴ [N2O]= 0.1x = 6.6 × 10–21‌mol‌L−1
© examsnet.com
Go to Question:
Prev Question
Next Question