Question

18. The equilibrium constant K, for the following reaction at 191°C is 1.24. What is K?
B(s) +-F2(8) 7 BF3(g)
3
2
For ti
(a) 6.7
(b) 0.61
(C) 8.30
(d) 7.6​

Answer 1

Explanation:

K

p

and K

c

are related as:

K

p

=K

c

(RT)

Δn

here, Δn=1−3/2=−1/2

Given: K

p

=1.24, R=0.082 Latm/K/mol, T=191+273=463K

Upon substitution we get:

1.24=K

c

. (0.082×463)

−1/2

K

c

=1.24×

0.082×463

K

c

=7.64

Therefore option D is correct.