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Question

The equlibrium constant K, for the reaction N2+3H22NH3 is 1.64×104atm at 400C . What will be the equilibrium constant at 500C, if heat of reaction in this temperature range is - 105185.8 Joules.



Correct Answer
A

0.144 × 10-4atm

Your Answer
B

4.144 × 10-3atm

Your Answer
C

0.144 × 10-2atm

Your Answer
D

0.144 × 10-1atm


Solution

The correct option is A

0.144 × 10-4atm


Kp1=1.64×104atm,Kp2=?

T1=400+273=673K

T2=500+273=773K

H=105185.8 Joules

R=8.314 J/K/mole

Applying equation

LogKp2=log1.64×104

105185.82.303×8.314(773673773×673)

or Kp2=0.144×104atm

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