Question

What is the half-life of A? What will the pressure, initially 32.1 kPa, at

(a) 10 s,
(b) 10 min after initiation of the reaction?

Answer 1

Answer:

a) 32.09 kPa

b) 32.09 kPa

Explanation:

Given data:

rate constant $= 3.56\times 10^{-7} s^{-1}$

initial pressure is = 32.1 kPa

half life of A is calculated as

$t_{1/2} = \frac{ln 2}{k}$

$t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}$

$t_{1/2} = = 1.956 \times 10^6 s$

for calculating pressure we have follwing expression

$ln p = ln P_o - kt$

$P =P_o e^{-kt}$

a) $P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa$

b) $P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa$