# 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'...

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

*print*Print*list*Cite

### 1 Answer

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

=>answer.

**Sources:**