Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C:
N2(g)+O2(g)→2NO(g)
7 answers
How, by guessing. There is no data.
delta Gf for NO(g) is 87.6 kJ/mol
delta Gf for both O2 and N2 gas is zero
delta Gf for both O2 and N2 gas is zero
Delta Gf= -RT*ln(K)
R is 8.314
T is 298
R is 8.314
T is 298
Running through I got the answer 1.97x10^-31 but it is telling me I am wrong.
e^(-87600/(8.314*298))= = 4.41128049e-16
check that, I had to use google as a calculator.
check that, I had to use google as a calculator.
I get 2*87.6*1000 = ? and
-?/8.314*298 = about -70 so
K = about 10^-70 or so. The number I obtained is 1.93E-71.
-?/8.314*298 = about -70 so
K = about 10^-70 or so. The number I obtained is 1.93E-71.
I just thought (about 10 hours after I posted this) I used log K and not ln K.
1 answer