D
the Keq expression of the 2nd one is = to (1/the first Keq expression)^(1/2)
so take the answer of the first Keq, 5.4x10^13 and put it in that equation
(1/(5.4x10^13))^(1/2)=1.36x10^-7=1.4x10^-7
The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C.
2NO (g) + O2 (g) ↔ 2NO2 (g)
What is the value of Keq at this temperature for the following reaction?
NO2 (g) ↔ NO (g) + 1/2 O2 (g)
The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C.
2NO (g) + O2 (g) ↔ 2NO2 (g)
What is the value of Keq at this temperature for the following reaction?
NO2 (g) ↔ NO (g) + 1/2 O2 (g)
a. 5.4 × 1013
b. 5.66 × 10-3
c. 5.4 × 10-13
d. 1.4 × 10-7
e. none of the above
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