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For reaction, PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)), K_c=1.79 "L mol"^(-1).Then at 500 K state the value of K_p with respect to R. |
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Answer» Solution :`K_p=K_c(RT)^(DELTAN)`, where `Deltan` =1+1-1=1 =1.79`[(R)(500)]^1` =895R |
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