Question

Consider the following reaction. N2(g) + O2(g) equilibrium reaction arrow 2 NO(g) If the equilibrium partial pressures of N2, O2, and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at 2200°C, what is KP?

Answer 1

The equilibrium constant, Kp, for the reaction N2(g) + O2(g) = 2 NO(g) at the given conditions is approximately 0.1010 when calculated using the given equilibrium partial pressures.

Step-by-step explanation:

The question is asking us to calculate the equilibrium constant, KP, for the reaction given the equilibrium partial pressures of the reactants and products at a certain temperature. Using the reaction N2(g) + O2(g) = 2 NO(g), we can express the equilibrium constant KP in terms of the partial pressures of the gases:

KP = (PNO)2 / (PN2 * PO2)

Substituting in the given partial pressures:

KP = (0.050 ATM)2 / (0.15 ATM * 0.33 ATM) = 0.0050 / 0.0495 = 0.1010

So, the equilibrium constant KP at 2200°C for this reaction is approximately 0.1010.

Answer 2

Step-by-step explanation:

As the given reaction equation is as follows.

`N_(2)(g) + O_(2)(g) \rightleftharpoons 2NO(g)`

Hence, expression for
`K_(p)` of this reaction is as follows.

`K_(p) = \frac{P^(2)_(NO)}{P_{N_(2)} * P_{O_(2)}}`

Now, putting the given values into the above expression as follows.

`K_(p) = \frac{P^(2)_(NO)}{P_{N_(2)} * P_{O_(2)}}`

=
`((0.050)^(2) atm)/((0.15 atm) * (0.33 atm))`

=
`5.05 * 10^(-2)`

Thus, we can conclude that the value of
`K_(p)` is
`5.05 * 10^(-2)`.