Question

Kp for reaction N2 + 3H2 EQUILIBRIUM with 2NH3 is 1.6 into 10 raise to minus 4/ATM square at 400 degree Celsius what will be KP at 500 degree Celsius heat of reaction in this temperature range is -25.14 kilocalorie

Answer 1

Answer:

1.2×10 raise to the power 4 atm

Answer 2

Answer:

Kp=1.53$X 10^{-4}$

Explanation:

Kp=Kc(RT$)^{n}$ where n is the no of moles of gaseous (products-reactants)

therefore Kc at 400C is given by:-

$\frac{Kp}{(RT)^{n}}$=$\frac{1.6 X10^{-4}}{(2(673))^{-2}}=1.6 X4X(673)(673)X10^{-4}$=289.8

log$(\frac{Kc}{Kc'} )$=$\frac{H}{2.303R}(\frac{T2-T1}{(T1)(T2)} )$

Kc=368

Now Kp' is given by

Kp=$(368)(2(773))^{-2}=\frac{368}{2390116}$≅1.53$X 10^{-4}$

Kp=1.53$X 10^{-4}$

NOTE: R value is 2 here (in calories...)

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