Question

For the reaction N2O4(g)⇌2NO2(g), ΔH0=+57.2kJmol−1 and KP=0.113 at 298 K

1) What is KP at 0∘C

2) At what temperature will KP= 1.00?

Answer 1

1) What is KP at 0∘C

Given that    K1= 0.113
K2=?
T1= 298 K
T2= 0∘C or 273 K
delta H= =+57.2kJmol−1 or 57200 J / mole

R= 8.314 (gas constant)

ln(K2/K1 )=dH /R (1/t1-1/ t2)

ln(x/0.113)=57200/8.314(1/298-1/273)

then;

ln(x/0.113)= -2.114
x/0.113= 0.1207.

to solve for x which is 0.014

2) At what temperature will KP= 1.00

K1= 1.00
K2=0.113
T1= x
T2= 298 K
delta H= 57200
R= 8.314 (gas constant)

ln(0.113/1.00)=57200/8.314(1/x-1/298)

-2.18=6880/x-23.087

X= 330 K

OR 57 c