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PCl5(g) <------> PCl3(g) + Cl2(g) if Kc of this reaction is 0.5 moldm^-3 in...
PCl5(g) <------> PCl3(g) + Cl2(g)
if Kc of this reaction is 0.5 moldm^-3 in 500K , then what is the Kp?
(I know the equation to slove this,but i have unit problem; that is how to get 'Pa' in the answer .So i need the answer in correct units) please help
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The balanced chemical reaction is:
PCl5 (g) → PCl3 (g) + Cl2 (g)
We know, `K_p = K_c * (RT)^(Deltan)`
Where ∆n is the net change in the number of moles of gaseous species in the reaction,
R is the Universal gas constant = 0.08205746 dm^3-atm./mol-K
Converting the pressure term into Pascal units (1 atm. = 101325 Pa),
R = 0.08205746*101325 dm^3-Pa/mol-K
= 8314.472 dm^3-Pa/mol-K
T is the temperature in the Absolute scale = 500 K
and ∆n = (1+1)-1 = 1
Putting the values in the relationship between Kp and Kc, we get
`K_p=0.5 * (8314.472*500)^1` [mol-dm^-3*dm^3-Pa*K/mol-K]
= 2078618 Pa
= 2078.618 kPa
Posted by llltkl on June 15, 2013 at 3:58 PM (Answer #1)
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