At a particular temperature Kp = 0.041 for the reaction N2(g) + O2(g) ⇄ 2NO(g) What...

At a particular temperature Kp = 0.041 for the reaction N2(g) + O2(g) ⇄ 2NO(g) What is the equilibrium partial pressure of NO if a flask initially contains 0.12 atm of all three gases? Report the pressure of NO in atm rounded to the nearest hundredth of an atm.

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Expert Solution

There's something I do not understand quite well, is the 0.12 atm the total of all gases, of that the innitial pressure for each gas?. If it's the total pressure, then we need more info for the other gases. If it's for each gas, then the procedure is the following:

r: N₂ + O₂ <------------> 2NO

i. 0.12 0.12 0.12

e. 0.12-x 0.12-x 0.12+2x

Kp = pNO² / pN₂ * pO₂

0.041 = (0.12+2x)² / (0.12-x)²

0.041(0.0144 - 0.24x + x²) = 0.0144 + 0.48x + 4x²

0.00059 - 0.00984x + 0.041x² = 0.0144 + 0.48x + 4x²

3.959x² + 0.48984x + 0.01381 = 0

Use the quadratic formula:

x = 0.4894 (0.4894² - 4*3.959*0.01381)^1/2 / 2*3.959

x = 0.4894 0.1443 / 7.918

x₁ = 0.044 atm

x₂ = 0.08 atm

The lowest value would be the one we'll take:

pNO = 0.12 + 2*0.044 = 0.208 atm

pN2 = pO2 = 012 - 0.044 = 0.076 atm

However confirm to me the value of atm that I stated at the beggining and tell me in a comment if there's something to be fixed.

Hope this helps

queen_honey_blossom answered 1 year ago

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