In: Chemistry
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.
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