Question

Practice Exercise 1 At 500 K, the reaction 2 NO(8) + Cl2(8) = 2 NOCI(8) has K, = 51. In an equilibrium mixture at 500 K, the

Practice Exercise 1 

At 500 K, the reaction 2 NO(8) + Cl2(8) = 2 NOCI(8) has K, = 51. In an equilibrium mixture at 500 K, the partial pres- sure of NO is 0.125 atm and Cl, is 0.165 atm. What is the par- tial pressure of NOCI in the equilibrium mixture? (a) 0.13 atm (b) 0.36 atm (c) 1.0 atm (d) 5.1 x 10-5 atm (e) 0.125 atm 


Practice Exercise 2

 At 500 K, the reaction PC15(8) = PC13(8) + Cl2(8) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCI, is 0.860 atm and that of PC13 is 0.350 atm. What is the partial pressure of Cl, in the equilibrium mixture?

0 0
Add a comment Improve this question Transcribed image text
Answer #1

~). The is to obj ective Nood in calculate the partial pressure Equilibrium mixture. of the Given data: kp= 56 partial partiaGiven pels (g) = pedz lg) + Cl2 lg ) 0.497 0.860 atm atm 0.350 partial kps partial pressure of pets pressure of pelz - [pos]

answered by: ANURANJAN SARSAM
Add a comment
Know the answer?
Add Answer to:
Practice Exercise 1 At 500 K, the reaction 2 NO(8) + Cl2(8) = 2 NOCI(8) has...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • At 500 K the reaction PCl5(g) <----> PCl3(g) + Cl2(g) has Kp = 0.497. In an...

    At 500 K the reaction PCl5(g) <----> PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?

  • A student ran the following reaction in the laboratory at 479 K PC13(E) + Cl2) PC...

    A student ran the following reaction in the laboratory at 479 K PC13(E) + Cl2) PC (e) When she introduced PCIy(R) and Cl() into a 1.00 L evacuated container, so that the initial partial pressure of PCI, was 2.39 atm and the initial partial pressure of Cl; was 1.94 atm, she found that the equilibrium partial pressure of Cl, was 0.373 am Calculate the equilibrium constant, Ky she obtained for this reaction.

  • The equilibrium constant, K, for the following reaction is 1.20x10-2 at 500 K. PC15(E) =PC13(E) +...

    The equilibrium constant, K, for the following reaction is 1.20x10-2 at 500 K. PC15(E) =PC13(E) + Cl2(E) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.157 M PCI3, 4.34x102 MPC12 and 4.34x102 M CI, What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.17x10 - mol of Cl2(g) is added to the flask? [PC13] = [PC13] = [Cl] =

  • The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g)...

    The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g) = PC15(g) Calculate the equilibrium concentrations of reactant and products when 0.389 moles of PC13 and 0.389 moles of Cl2 are introduced into a 1.00 L vessel at 500 K [PC13] = [Cl] = [PC15] - The equilibrium constant, Kc, for the following reaction is 5.10x10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibrium concentration of HCl when 0.467 moles of NHACI(S)...

  • The equilibrium constant, K., for the following reaction is 1.20 10-2 at 500 K. PCI(8) =PC13(g)...

    The equilibrium constant, K., for the following reaction is 1.20 10-2 at 500 K. PCI(8) =PC13(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.363 moles of PC13() are introduced into a 1.00 L vessel at 500 K M [PC131 = [PC13] - [Cl] M M

  • Consider the following reaction where K. = 83.3 at 500 K: PC13(g) + Cl2(g) = PC15(8)...

    Consider the following reaction where K. = 83.3 at 500 K: PC13(g) + Cl2(g) = PC15(8) A reaction mixture was found to contain 2.76x10-2 moles of PC13(g), 4.46x10-2 moles of Cl2(g) and 0.133 moles of PC13(g), in a 1.00 Liter container. Indicate True (T) or False (F) for each of the following: (- 1. In order to reach equilibrium PC15(g) must be consumed. 2. In order to reach equilibrium K, must increase. 3. In order to reach equilibrium PC1z must...

  • The squares in the equation are equilibrium signs. The equilibrium constant, Ky, for the following reaction...

    The squares in the equation are equilibrium signs. The equilibrium constant, Ky, for the following reaction is 0.110 at 298 K: NH_HS(s) NH3(g) + H2S(g) Calculate the equilibrium partial pressure of H2S when 0.416 moles of NH_HS(s) is introduced into a 1.00 L vessel at 298 K. Phys= c atm The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PC15(g) PC13(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PC15(g) is introduced into...

  • Consider the following reaction where Kc = 1.20x10-2 at 500 K: PCl; (g)PCl3 (g) + Cl2...

    Consider the following reaction where Kc = 1.20x10-2 at 500 K: PCl; (g)PCl3 (g) + Cl2 (g) A reaction mixture was found to contain 0.125 moles of PCls (2),4.99-102 moles of PCl3 (g), and 4.1 Indicate True (T) or False (F) for each of the follow ー▼ I. In order to reach equilibrium PC15(g) must be produced in a ▼ 2. In order to reach equilibrium Kc must decrease ▼ 3. In order to reach equilibrium PC13 must be consumed...

  • The equilibrium constant, K, for the following reaction is 3.52*10 at 528 K PC13(g) =PC12(g) +...

    The equilibrium constant, K, for the following reaction is 3.52*10 at 528 K PC13(g) =PC12(g) + Cl2(g) An equilibrium mixture of the three gases in a 9.42 L container at 528 K contains 0.244 M PCIE. 9.27X10-2M PCI, and 9.27-10-2M Cly. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 17.1 L? M [PC15] [PC13] = [Cl] M M

  • Question 1. Phosphorous trichloride reacts with chlorine to produce phosphorus pentachloride: PC13(g) + Cl2(8) - PC1s(8)...

    Question 1. Phosphorous trichloride reacts with chlorine to produce phosphorus pentachloride: PC13(g) + Cl2(8) - PC1s(8) The equilibrium constant (Kc) for the reaction is 96 at 400 K If the equilibrium concentration of PC13 is 0.50 M and Cl, is 0.070 M, what is the equilibrium concentration of PCI ? Question 2. Consider the reaction between hydrogen and iodine H2(g) + 12(6) 2 HI(g) Kc = 64 Initially, a container was charged with 0.55 atm of H, and I2, what...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT