2NO (g) + Cl_2 (g) rightwardsarrowoverleftwardsarrow 2NOCl (g) K_p = 1.4 times 1

Question

2NO (g) + Cl_2 (g) rightwardsarrowoverleftwardsarrow 2NOCl (g) K_p = 1.4 times 10^5 Since K_p is large, the reaction goes close to completion. In such cases, first react to completion, and then evaluate. In this case, the limiting reactant is Cl_2. That will get close to 0, forming NOCl and leaving over excess NO 2NO (g) + Cl_2 (g) rightwardsarrowoverleftwardsarrow 2NOCl (g) NO and NOCl pressures are 0.30, 0.20 respectively. The Cl_2 is very small but not quite zero. We can solve for the Cl_2 pressure (represented by x) by just substituting into the K_p expression. K_p = (0.20)^2/(0.30)^2 x = 1.4 times 10^5 Solving: x = 3.2 times 10^-6 atm Choice E

Explanation / Answer

Partial pressure of the gas is proportional to its concentration . A limiting agent will have lower partial pressure than the reagent present in excess. You can substitute pressure for concentration.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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