1.

A first order gas reaction A_(2)B_(2)(g) to 2A (g) +2B(g) at the temperature 400^(@)C has the rate constantk=2.0xx10^(-4)s^(-1). What percentage of A_(2)B_(2) is decomposed on heatingfor 900 seconds?

Answer»

Solution :For the first ORDER reaction, use the relation
`k=(2.303)/(t)"log"([R]_(0))/([R])`
Given that `k=2.0xx10^(-4)s^(-1) and t =900s`
Substituting the values in the first order equation, we have
`2.0xx10^(-4) =(2.303)/(900) "log" ([R]_(0))/([R])`
or `"log"([R]_(0))/([R])=(2.0xx10^(-4)xx900)/(2.303)=0.0781`
TAKING antilogarithm of the above
`([R]_(0))/([R])=1.197 or ([R])/([R]_(0))=(1)/(1.197)=0.8354`
or Fraction decomposed `= 1-0.8354=0.1646`
Percentageof `A_(2)B_(2)` decomposed `=0.1646xx100=16.46`.


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