Question

The decomposition of sulfuryl chloride (so2cl2) is a first-order process. the rate constant for the decomposition at 660 k is 4.5 ✕ 10-2 s-1. (a) if we begin with an initial so2cl2 pressure of 330. torr, what is the pressure of this substance after 73 s? torr (b) at what time will the pressure of so2cl2 decline to one half its initial value? s

Answer 1

THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION : lnPso2cl12=-kt+lnPso2cl1 initial partial pressure Pso2cl12 the rate constant k and the time t lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375) so lnPso2cl12=3.002 we take the base e antilog: lnPso2cl12=e3.002 Pso2cl12=20 torr we use the integrated first order rate lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002 we use the same rate constant and initial pressure k=4.5*10-2*s-1 Pso2cl12=375 Pso2cl12=1* so2cl12 Pso2cl12=37.5 torr subtract in Pso2cl12 grom both side lnPso2cl12- lnPso2cl12=-kt ln(x)-ln(y)=ln (x/y) ln (Pso2cl12/Pso2cl20)=-kt we get t -1/k*ln(Pso2cl12/Pso2cl20)=t t=51 s