A student ran the following reaction in the laboratory at 446 K: PCl5(g) PCl3(g)

Question

A student ran the following reaction in the laboratory at 446 K: PCl5(g) PCl3(g) + Cl2(g) When she introduced 0.929 moles of PCl5(g) into a 1.00 Liter container, she found the equilibrium concentration of Cl2(g) to be 3.07E-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction The equilibrium constant, Kc, for the following reaction is 1.21 at 658 K. 2 NH3(g) N2(g) + 3 H2(g) When a sufficiently large sample of NH3(g) is introduced into an evacuated vessel at 658 K, the equilibrium concentration of H2(g) is found to be 0.870 M. Calculate the concentration of NH3 in the equilibrium mixture.

Explanation / Answer

Kc = [Cl2][PCl3]/[PCl5]

at equi [PCl5] = ( 0.929-x)/1 , [PCl3]=[Cl2] =x = 0.0307/1 ( given)

hence [PCl5] equi = ( 0.929-0.0307) = 0.8983/1

now K c = ( 0.0307^2) /( 0.8983) = 0.00105 = 1.05 x 10^-3

2) equi [H2] = 0.87 hence equi [N2] = 0.87/3 = 0.29 M

Kc = [N2][H2]^3/[NH3]^2

1.21 = ( 0.29)(0.87)^3/[NH3]

[NH3] equi = 0.158 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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