Question

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.46×10−4 s−1 at a certain temperature.

1) What is the half-life for this reaction?

2) How long will it take for the concentration of SO2Cl2 to decrease to 25% of its initial concentration? Express your answer using two significant figures.

3) If the initial concentration of SO2Cl2 is 1.00 M, how long will it take for the concentration to decrease to 0.80 M ?

4) If the initial concentration of SO2Cl2 is 0.150 M , what is the concentration of SO2Cl2 after 220 s ?

5) If the initial concentration of SO2Cl2 is 0.150 M , what is the concentration of SO2Cl2 after 450 s ?

Answer 1

constant k is given by

-ln (1-XA)= kt      (1)

1) For calculating half life , put XA=0.5 ( time required for 50% drop in concentration is known as half life)

Which gives half life= 0.693/k= 0.693/1.46*10^-4 =4746.6 sec

2) XA= 1-CA/CAO, where CA= Concentration at any time t and CAO= initial concentration

it is given that CA=0.25CAO and hence XA=1-0.25=0.75

from (1) –ln(1-0.75)= 1.46*10^-4*t

therefore time, t= 9495.2 seconds

3) given CA =1M and CAO= 0.8M    XA=1-0.8= 0.2

From (1) –ln (1-0.2)= 1.46*10^-4*t ,t= 1528.38 sec

4) CAO= 0.15M t= 220 sec

From (1)= -ln(1-XA)= 1.46*10^-4*220=0.03212 , 1-XA= exp(-0.0312)= 0.96839

CA/CAO= 0.96389 and CA= 0.96389*0.15=0.1453M

5) given CAO= 0.15M and t= 450 sec

From (1) –ln(1-XA)= 1.46*10^-4*450=0.0657 and 1-XA=CA/CAO= 0.9364 and CA=0.9364*0.15=0.140M