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the equilibrium constant kp for the following reaction is found to be 4.31 *10^-4 at 375...

the equilibrium constant kp for the following reaction is found to be 4.31 *10^-4 at 375 degrees C. N2(g)+3H2(g)<-->2NH3(g) In a certain experiment, a student starts with .862 atm N2 and .373 atm of H2 in a constant volume vessel at 375 degrees celcius. Which of the following would be the closest to the equilibrium partial pressure of NH3? Do I use the mole fraction to find Partial Pressure?

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Answer #1

No need to use mole fraction

N2(g)         +         3H2(g)<-->         2NH3(g)

0.862-x                    0.373-3x             2x

Kp= (pNH3)^2/ (pN2)(pH2)^3

4.31 *10^-4 = (2x)^2/ (0.862-x)(0.373-3x)^3

solve this to get pNH3= 2x = 4.4 * 10^-3 atm

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