The decomposition of phosphine, PH_(3), proceeds according to the following equation: 4PH_(3)(g) to P_(4)(g) +6H_(2)(g) It is found that the reaction follows the following rate equation: Rate =k[PH_(3)]. The half-life of PH_(3) is 37.9s at 120^(@)C. (i) How much time is required for 3//4th of PH_(3) to decompose? (ii) What fraction of the original sample of PH_(3) remains behind after 1 minute?

The decomposition of phosphine, PH_(3), proceeds according to the foll

1.	The decomposition of phosphine, PH_(3), proceeds according to the following equation: 4PH_(3)(g) to P_(4)(g) +6H_(2)(g) It is found that the reaction follows the following rate equation: Rate =k[PH_(3)]. The half-life of PH_(3) is 37.9s at 120^(@)C. (i) How much time is required for 3//4th of PH_(3) to decompose? (ii) What fraction of the original sample of PH_(3) remains behind after 1 minute?
Answer» Solution :`4PH_(3)(g) to P_(4)(g) +6H_(2)(g)` Half-life of `PH_(3)=37.9s` Rate `=k[PH_(3)]` This MEANS that the reaction is of first order. For a first order reaction `k=(0.693)/(t_(1//2)) or k=(0.693)/(37.9)=0.01828496` (i) Apply the relation `k=(2.303)/(t) "log"([R]_(0))/(R )` Here `R=R_(0)-(3)/(4)R_(0)=(R_(0))/(4)` Substituting the values in the equation above, we have `0.01828496=(2.303)/(t)"log"([R]_(0))/([R_(0)]//4)` or `t=(2.303)/(0.01828496)log 4 or t=(2.303)/(0.01828496)xx 0.60206=75.83s` (ii) After 1 minute or 60 seconds `k=(2.303)/(t)"log"([R]_(0))/(R ) or " log"([R]_(0))/(R )=(kt)/(2.303)` Substituting the values, we have `"log"([R]_(0))/(R )=(0.0183xx60)/(2.303)=0.4768` Taking antilogarithm, we have `([R]_(0))/(R )=2.992 or (R )/([R]_(0))=(1)/(2.992)=0.3342`