Answer:
a) 32.09 kPa
b) 32.09 kPa
Explanation:
Given data:
rate constant [tex]= 3.56\times 10^{-7} s^{-1}[/tex]
initial pressure is = 32.1 kPa
half life of A is calculated as
[tex]t_{1/2} = \frac{ln 2}{k}[/tex]
[tex]t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}[/tex]
[tex]t_{1/2} = = 1.956 \times 10^6 s[/tex]
for calculating pressure we have follwing expression
[tex]ln p = ln P_o - kt[/tex]
[tex]P =P_o e^{-kt}[/tex]
a) [tex]P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa[/tex]
b) [tex]P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa[/tex]
