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For the reaction PCl3(g)+ Cl2(g)= PCl5(g), the value of Kc at 250 degree celsius is 26. Then value of Kp at this temperature is(a) 0.61 (b) 0.57(c) 0.83 (d) 0.46

For the reaction PCl3(g)+ Cl2(g)= PCl5(g), the value of Kc at 250 degree celsius is 26. Then value of Kp at this temperature is(a) 0.61 (b) 0.57(c) 0.83 (d) 0.46

Grade:12th pass

2 Answers

Shivam
19 Points
6 years ago
(a) 0.61Kp = Kc× (RT)^∆n∆n=(no of moles of gaseous products - no of moles of gaseous reactants)Therefore, ∆n = 1-2= (-1)Now, Kc=26, R=0.0821, T= 250+273.15=523.15KHence, Kp= 26×(RT)^ -1=26/RT = 26/(0.0821×523.15) = 0.61
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached solution to the problem.

Kp = Kc×(RT)∆ng
∆ng = no of moles of gaseous products - no of moles of gaseous reactants
Therefore, ∆ng = 1-2 = (-1)
Now, Kc = 26, R = 0.0821, T = 250+273.15 = 523.15K
Hence, Kp = 26×(RT)-1 = 26/RT = 26/(0.0821×523.15) = 0.61
Hence a) is the correct option.
Hope it helps.
 
Thanks and regards,
Kushagra
 

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