Chemistry

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

asked by DrewS
  1. 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

    posted by Amanda

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