Question

A gaseous reaction A(g) $\leftrightarrow$ 2B(g) + C(g) is found to be of first order. If the reaction is started with p_A = 90 mm Hg, the total pressure after 10 min is found to be 180 mm Hg. The rate constant of the reaction is

• A. $4.60\times {10}^{-3}{s}^{-1}$
• B. $6.9\times {10}^{-2}{s}^{-1}$
• C. $1.15\times {10}^{-3}{s}^{-1}$
• D. $4.90\times {10}^{-3}{s}^{-1}$

Answer 1

The correct option is C $1.15\times {10}^{-3}{s}^{-1}$

A(g) $\to$ 2B(g) + C(g)
$p_0$ 0 0
After 10 min : p 2p p
Total pressure after 10 min = $p_0$ + p + 2p + p
180mm = 90mm + 2p
or, 2p = 90mm Hg
p = 45mm Hg
$Now,k=\frac{2.303}{t}log\frac{p_0}{p}=\frac{2.303}{10min}log\frac{p_0}{p}=\frac{2.303}{10min}log\frac{90}{45}=\frac{2.303}{10min}\times 0.303=1.155\times {10}^{-3}{s}^{-1}$