At high temperatures elemental nitrogen and oxygen react with each other to form

Question

At high temperatures elemental nitrogen and oxygen react with each other to form nitrogen monoxide: N_2 (g) + O_2 (g) doubleheadarrow 2 NO (g) Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be [N_2] = 0.041 M. [O_2] = 0.0078 M and [NO] = 4.7 times 10^-4 M. Calculate K for the reaction. At a Particular temperature, a 2.00 L flask at equilibrium contains 2.80 times 10^-4 mol of N_2, 2.50 times 10^-5 mol of O_2, and 2.00 times 10^-2 mol of N_2O. Calculate K at this temperature for the reaction: 2N_2(g) + O_2(g) doubleheadarrow 2 N_2O(g

Explanation / Answer

1)

k = [NO]^2 / [N2][O2]

= (4.7*10^-4)^2 / (0.041 * 0.0078)

= 6.91*10^-4

Answer: 6.91*10^-4

2)

[N2] = number of mol / volume

= (2.80*10^-4 mol) / 2.00 L

= 1.40*10^-4 M

[O2] = number of mol / volume

= (2.50*10^-5 mol) / 2.00 L

= 1.25*10^-5 M

[N2O] = number of mol / volume

= (2.00*10^-2 mol) / 2.00 L

= 1.00*10^-2 M

Kc = [N2O]^2 / [N2]^2 [O2]

= (1.00*10^-2)^2 / (1.40*10^-4)^2 * (1.25*10^-5)

= 4.08*10^8

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