Question

Consider a reaction A(g) ¾® 3B(g) + 2C(g) with rate constant is 1.386 × 10^–2 min^–1 in a non-rigid closed container starting with 2 moles of A in 12.5 lit vessel initially, if reaction is allowed to take place at constant pressure & at 298 K the conc. of B after 100 min is -

• A. 0.18 M
• B. 0.03 M
• C. 0.09 M
• D. 0.01 M

Answer 1

Answer: (C)

0.09 M

Answer 2

A ¾® 3B + 2C

K = $\frac{2.303}{t}$log $\frac{\lbrack A\rbrack_{0}}{\lbrack A\rbrack}$

Ž [A] = 0.0016

Moles of A decomposed = 1.98

So moles of B formed = 5.94

Assuming n µ v final vol = 62 lit

So find conc. of B = $\frac{5.94}{62}$ = 0.09 mol/lit