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Find out the value of K_c for each of the following equilibria from the value of K_p : (i)2NOCl_((g)) hArr 2NO_((g)) + Cl_(2(g)) , K_p=1.8xx10^(-2) , 500 K (ii)CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)), K_p=167 , 1073 K |
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Answer» Solution :(i)`{:("Reaction equilibrium:",2NOCl_((g)) hArr, 2NO_((g))+, Cl_(2(g))),("Stoichiometric multiply:", 2,2,1):}` All are gaseous , So, `Deltan_((g))` = (Difference of coefficient of products and reactant gas compount) `therefore Deltan_((g))`=(2+1)-(2)=+1 T=500 K, `K_p=1.8xx10^(-2)` R=0.0831 L bar `"mol"^(-1) K^(-1)` `K_p=K_c(RT)^(DELTAN)_((g))` `therefore K_c=K_p/(RT)^(Deltan_((g)))=(1.8xx10^(-2))/{{(0.0831)(500)}}^1` `=(1.8xx10^(-2))/(0.0831xx5xx10^2)` `=4.3321xx10^(-4) "mol L"^(-1)` (If R=0.0821 L atm `"mol"^(-1) K^(-1)` So, answer `4.384xx10^(-4) "mol L"^(-1)` ) (ii)`CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))` In equilibrium SOLID is ignore, `Deltan_((g))`=(Coefficient of gaseous product )-(Coefficientof gaseous reactants ) =+1-0=+1 `K_p`=167 T=1073 K R=0.0831 L bar `"mol"^(-1) K^(-1)` |
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