A) more than one
B) less than one
C) equal to \[{{K}_{c}}\]
D) zero
Correct Answer: C
Solution :
Key Idea The relation between \[{{K}_{p}}\] and \[{{K}_{c}}\] is given by \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] where \[\Delta \,{{n}_{(g)}}\] number of moles of products - number of moles of reactants For the following equation \[{{H}_{2}}(g)+{{I}_{2}}(g)2HI(g)\] \[\Delta {{n}_{g}}=2-2=0\] \[{{K}_{p}}={{K}_{c}}\,{{(RT)}^{0}}\] \[{{K}_{p}}={{K}_{c}}\] Note - Fore the reaction \[{{n}_{1}}A+{{n}_{2}}B{{m}_{1}}C+{{m}_{2}}D\] the following cases are possible 1. When \[({{m}_{1}}+{{m}_{2}})>({{n}_{1}}+{{n}_{2}})\] eg, \[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\] then \[{{K}_{p}}>{{K}_{c}}\] 2. When \[({{m}_{1}}+{{m}_{2}})<({{n}_{1}}+{{n}_{2}})\] eg. \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}}\] then \[{{K}_{p}}<{{K}_{c}}\] 3. When \[({{m}_{1}}+{{m}_{2}})=({{n}_{1}}+{{n}_{2}})\] eg, \[{{H}_{2}}+C{{l}_{2}}2HCl\] then \[{{k}_{p}}={{k}_{c}}\]
You need to login to perform this action.
You will be redirected in
3 sec
