The equilibrium constant Kp for the following reaction is 4.31x10^-4 at 375 cels

Question

The equilibrium constant Kp for the following reaction is 4.31x10^-4 at 375 celsius: N2(g) + 3H2(g) <-> 2NH3(g) In a certain experiment a student starts with 0.862 atm of N2 and 0.373 atm of H2 in a constant volume vessel at 375 celsius. Calculate the partial pressures of all species when equilibrium is reached. The equilibrium constant Kp for the following reaction is 4.31x10^-4 at 375 celsius: N2(g) + 3H2(g) <-> 2NH3(g) In a certain experiment a student starts with 0.862 atm of N2 and 0.373 atm of H2 in a constant volume vessel at 375 celsius. Calculate the partial pressures of all species when equilibrium is reached. In a certain experiment a student starts with 0.862 atm of N2 and 0.373 atm of H2 in a constant volume vessel at 375 celsius. Calculate the partial pressures of all species when equilibrium is reached.

Explanation / Answer

Given -

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

I(atm)     0.862       0.373           0

C              -x           -3x              +2x

Eq        (0.862-x)    (0.373-3x)     2x

Since - Kp = PNH32 / PH23 * PN2

4.31x10^-4 = (2x)2 / (0.862-x)(0.373-3x)3

By solving, x = 0.00219 atm

Thus, Partial pressure of N2 = 0.862 -0.00219 = 0.859 atm

Partial pressure of H2 = 0.366 atm

Partial pressure of NH3 = 2(0.00219) = 0.00438 atm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.