Consider the following equilibrium, for which K_p = 7.54×10−2?
Consider the following equilibrium, for which K_p = 7.54×10−2 at 460 degree celsius. 2Cl2(g)+2H2o(g)-->(equilibrium)4Hcl(g)+O…
A)What is the value of K_p for the reaction
4Hcl=O2-->2Cl2+2H2O?
B) What is the value of K_p for the reaction
2Cl2(g)+H2o(g)-->(equilibrium)2Hcl(g)+1/…
C) What is the value of K_c for the reaction in part B?
please help me understnad it . Thx!!
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Verified answer
Given:
2 Cl2 + 2 H2O <=> 4 HCl + O2
Kp=p(HCl)^4*p(O2)/[p(Cl2)²*p(H2O)²]
a)
4 HCl + O2 <=> 2 Cl2 +2 H2O
K= p(Cl2)²*p(H2O)²/[p(HCL)^4*p(O2) ]
Compare K to Kp and find out that K=1/Kp
b) Sorry, the stoichometry of your reaction isn't correct. Is the reaction
Cl2 + H2O <=> 2 HCl + O ? ( molecular oxygen isn't stable!)
Then K= p(HCl)² p(O)([p(Cl2)*p(H2O)]
Comparing K to Kp:
p(Cl2)* p(H2O)= p(HCl)²* sqrt(p(O2))/ sqrt(Kp) // sqrt: square root
-> K= p(O)*sqrt(Kp)/sqrt(p(O2)))
c) Convert every pressure to concentration by assuming ideal gas behavior. C=P/RT and Kp becomes Kp= c(HCl)²*c(O)* (RT)^-3/[c(Cl2)*c(H20)* (RT)^ -2) ]
Kp=Kc* (RT)^-1