Question

The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g)...

The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K:

CH4(g) + CCl4(g) <---> 2CH2Cl2(g)

Calculate the equilibrium concentrations of reactants and product when 0.343 moles of CH4 and 0.343 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.

[CH4] = M
[CCl4] = M
[CH2Cl2] = M
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Answer:-

This question is solved by using simple concept of chemical equilibrium. Firstly equilibrium constant expression is written then using the its value equilibrium concentration is calculated.

The answer is given in the image,

Add a comment
Know the answer?
Add Answer to:
The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g)...
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
  • a) The equilibrium constant, Kc, for the following reaction is 9.52×10-2at 350 K. CH4(g) + CCl4(g)...

    a) The equilibrium constant, Kc, for the following reaction is 9.52×10-2at 350 K. CH4(g) + CCl4(g) -> 2 CH2Cl2(g) Calculate the equilibrium concentrations of reactants and product when 0.251 moles ofCH4and 0.251 moles of CCl4are introduced into a 1.00 L vessel at 350 K. [ CH4] = M [ CCl4] = M [ CH2Cl2] = M b) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) ->PCl3(g) + Cl2(g)   Calculate the equilibrium concentrations of reactant...

  • The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) >>CH4(g) +...

    The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) >>CH4(g) + CCl4(g) Calculate the equilibrium concentrations of reactant and products when 0.327 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K. [CH2Cl2] = M [CH4] = M [CCl4] = M The equilibrium constant, Kc, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) >>CO(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.275 moles of COCl2(g) are...

  • A) The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g)   CH4(g)...

    A) The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g)   CH4(g) + CCl4(g)   Calculate the equilibrium concentrations of reactant and products when 0.391 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K.   [CH2Cl2] = M [CH4] = M [CCl4] = M B) The equilibrium constant, Kc, for the following reaction is 1.80×10-4 at 298 K. NH4HS(s) NH3(g) + H2S(g)   Calculate the equilibrium concentration of H2S when 0.260 moles of NH4HS(s) are...

  • The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) = CH4(g)...

    The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) = CH4(g) + CCl4(g) Calculate the equilibrium concentrations of reactant and products when 0.294 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K. [CH2Cl2] = ? M [CH4] = ? M [CCl4] = ? M

  • 34) The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) <--->...

    34) The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) <---> CH4(g) + CCl4(g)   Calculate the equilibrium concentrations of reactant and products when 0.222 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350K.   [CH2Cl2] = ___ M [CH4] = ___ M [CCl4] = ___ M

  • The equilibrium constant, Kc, for the following reaction is 9.52x10-2 at 350 K: CH4(8) + CC14(8)...

    The equilibrium constant, Kc, for the following reaction is 9.52x10-2 at 350 K: CH4(8) + CC14(8) — 2CH2Cl2(g) Calculate the equilibrium concentrations of reactants and product when 0.305 moles of CH4 and 0.305 moles of CCl4 are introduced into a 1.00 L vessel at 350 K. [CH] = [CC14] = [CH2Cl2] = A student ran the following reaction in the laboratory at 531 K: COC12(8) CO(g) + Cl2(8) When she introduced 1.87 moles of COCl2(g) into a 1.00 liter container,...

  • The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) = CH4(g)...

    The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) = CH4(g) + CC14(9) Calculate the equilibrium concentrations of reactant and products when 0.377 moles of CH,Cl2 are introduced into a 1.00 L vessel at 350 K (CH2Cl2] = [CH4] = [CCl4] =

  • The equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g)...

    The equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g) --> 2CH2Cl2(g) Calculate the equilibrium partial pressures of all species when CH4 and CCl4, each at an intitial partial pressure of 0.959atm, are introduced into an evacuated vessel at 350 K.   PCH4 = atm PCCl4 = atm PCH2Cl2 = atm

  • 1. The equilibrium constant, Kc, for the following reaction is 42.2 at 288 K. 2CH2Cl2(g) <----->...

    1. The equilibrium constant, Kc, for the following reaction is 42.2 at 288 K. 2CH2Cl2(g) <-----> CH4(g) + CCl4(g)   When a sufficiently large sample of CH2Cl2(g) is introduced into an evacuated vessel at 288 K, the equilibrium concentration of CCl4(g) is found to be 0.140 M.   Calculate the concentration of CH2Cl2 in the equilibrium mixture. _____ M 2. A student ran the following reaction in the laboratory at 295 K: 2CH2Cl2(g) <-------> CH4(g) + CCl4(g)   When she introduced 8.46×10-2 moles...

  • The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(8) CH4(8) +...

    The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(8) CH4(8) + CC14(8) Calculate the equilibrium concentrations of reactant and products when 0.242 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K. M (CH2Cl2] [CH] [CC14] M M

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