1.

PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K in a closed container and their concentrations are 0.8xx10^(-3) mol L^(-1), 1.2xx10^(-3)mol L^(-1) respectively. The value of K_(c ) for the reactionPCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)will be

Answer»

`1.8xx10^(-3) mol L^(-1)`
`1.8xx10^(3)`
`1.8xx10^(-3)L mol^(-1)`
`0.55xx10^(4)`

Solution :`K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/((0.8xx10^(-3)))`
`=1.8xx10^(-3) mol L^(-1)`.


Discussion

No Comment Found

Related InterviewSolutions