a) Calculate K at 298 K for
the following reaction:
NO(g) +
1 |
2 |
O2(g) ? NO2(g)
?G0f(kj/mol) |
|||||
NO(g) | 86.60 | ||||
NO2(g) | 51 | ||||
O2(g) | 0 |
K = ×10
ANSWER:
we know that,
?G° = ? ?Gf°(products)– ? ?Gf°(reactants)
?G° = {51.0 - (86.60 + 0.0)} kJ/mol
?G° = - 35.60 kJ/mol = 3.56 x 104 J/mol
and, relation between ?G° and K is:
?G° = - 2.303 RT log K
-3.56 x 104 J/mol = - 2.303 x 8.314 J/K-mol x 298K x log K
log K = 6.24
K = 1.74 x 106
Hence, K at 298 K for the following reaction is 1.74 x 106.
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