1.

PCl_(5),PCl_(3) "and "Cl_2 are at eqilibrium at 500 K and having concentration 1.59 M PCl_(3), 1. 59 M Cl_2 and 1.41 M PCl_5 Calculate K_C for the reaction PCl_5 hArr PCl_3 +Cl_2

Answer»

Solution :The equlibrium CONSTANT `K_C` for the above reaction can be WRITTENAS : `K_(C)=([PCl_3][Cl_2])/([PCl_5])=((1.59)^2)/(1.41)=1.79`


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