log kp/kc + logrt =0
Answers
Answer:
Hey!
For the general reaction, aA+bB⇋cC+dD
the relationship between two equilibrium constants is: K
P
=K
C
(RT)
Δn
where, Δn (Total moles of products side) - (Total moles of reactants on the reactant side). Hence Δn= (d+c) - (a+b). R is the gas constant found in the ideal gas law (0.0821 liter x Atm/mole/kelvin) T is the temperature of reaction, kelvin.
This we can use in a relationship for the reaction:
log
K
C
K
P
+logRT=0
log
K
C
K
C
(RT)
Δn
+logRT=0
log((RT)
Δn
×(RT)=0
log(RT)
Δn+1
=0
(RT)
Δn+1
=0
R is a constant and T is the temperature of reaction, so the product cant be zero. Thats why Δn+1=0 and Δn=−1
so our relationship is true for the reaction (B) 2SO
2
+O
2
⇋2SO
3
Because Δn for this reaction is
Δn =2-2-1= -1
logkp/kc + logrt = 0
logkp *rt/kc = log 1
kp*rt/kc = 1
kp*rt = kc
Formula used
log A + log b = logA* b
log 1 = 0