If kc = 4.0×10−2 for pcl3(g)+cl2(g)⇌pcl5(g) at 490 k , what is the value of kp for this reaction at this temperature? express your answer using two significant figures. g

The given reaction is :

PCl3 (g) + Cl2(g) ↔ PCl5 (g)

The relationship between the equilibrium constants, Kc (based on gas concentration) and Kp(based on gas partial pressure) is:

Kp = Kc * (RT)^Δn (1)

where : Δn = change in the number of moles = n(products) - n(reactants)

R = gas constant = 0.0821 L-atm/mol-K

T = temperature

for the given reaction:

Δn = 1-(1+1) = -1

T = 490 K

Kc = 4.0 * 10^-2

Based on eq (1):

Kp = 4.0*10^-2 *(0.0821* 490)^-1 = 9.9 * 10^-4

that's just how they are

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the answer is b. dilution