The equilibrium constant K_P for the reaction PCl_5(g) rightarrow PCl_5(g) + Cl_

Question

The equilibrium constant K_P for the reaction PCl_5(g) rightarrow PCl_5(g) + Cl_2(g) is 1.05 at 250 degree C. The reaction starts with a mixture of PCl_5, PCl_3, and Cl_2 at pressures 0.177 atm, 0.223 atm, and 0.111 atm, respectively, at 250 degree C. When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? A 2.50-mole quantity of NOC1 was initially in a 1.50-L reaction chamber at 400 degree C. After equilibrium was established, it was found that 28.0 percent of the NOC1 had dissociated: 2NOCl(g) rightarrow 2NO(g) + Cl_2(g) Calculate the equilibrium constant K_c for the reaction. Consider the following reaction, which takes place in a single elementary step: 2A + B A_2B If the equilibrium constant K_c is 12.6 at a certain temperature and if k_r = 5.1 Times 10^-2 s^-1, calculate the value of k_f. the units for k_f would be M^-2s^-1.

Explanation / Answer

8.   

pcl5 <===> pcl3 + cl2

Kp = pPCl3*pCl2 / pPCl5 = 1.05

q = (0.223*0.111)/0.177 = 0.14

as q>kp the reaction procedds towards right.so that concentration of pCl3,cl2 decreases , pcl5 increases.

8.

8. 2 NOCl <====> 2NO + Cl2

kc = [NO]^2[Cl2]/[NOCl]^2

concentration of NOCl = 2.5/1.5 = 1.67 M

reaction percentage = 1.67*28/100 = 0.4676

at equilibrium

[NOCl] = 1.67-0.4676 = 1.2024 M

[NO] = 2*0.4676 = 0.9352 M

[Cl2] = 0.4676 M

Kc = (0.9352^2*0.4676 / 1.2024^2)

   = 0.283

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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