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Which type of relation of K_p and K_c when Deltan=0, Deltan gt 0 and Deltan lt 0 |
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Answer» Solution :If `Deltan=0`, So `K_p=K_c` Its mean, total no. of moles of products and reactants are equal e.g.: `H_(2(g)) + I_(2(g)) hArr 2HI_((g)) , K_p= K_c` If `Deltan gt 0` so, (c+d) `gt` (a+b) and `K_p gt K_c` Its mean , total no. of moles of products are more than reactants so, `K_p gt K_c` and `Deltan` = positive no. e.g.: `PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))` `Deltan`= Mole of product - Mole of REACTANT =2-1=+1 So, `K_p gt K_c` If `Deltan lt 0`, (c+d) `lt` (a+b) and `K_p lt K_c` Its mean, total no. of moles of products are less than reactants so , `K_p lt K_c` and `Deltan`=negative no. e.g. : `{:(N_(2(g))+ , 3H_(2(g)) hArr , 2NH_(3(g))),(1,3,2):}` For (a+b)=(1+3)=4 and for product (c+d)=2 `Deltan`=(c+d)-(a+b)=2-4=-1 `Deltan`=negative no. `THEREFORE Deltan lt 0` `therefore K_p lt K_c` for this reaction. |
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