Suppose a 250. mL flask is filled with 1.8 mol of Cl2, 1.5 mol of HCl and 0.10 mol of CC14. The following reaction becomes possible:Cl₂(g) + CHCl3 (g) →HC1 (g) +CC14 (8)The equilibrium constant K for this reaction is 0.642 at the temperature of the flask.Calculate the equilibrium molarity of CHC13. Round your answer to two decimal places.MXŚ

Volume of the flask = 250ml Hence initial concentration of Cl2 since from the…

Answered: Suppose a 250. mL flask is filled with…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Suppose a 250. mL flask is filled with 1.8 mol of Cl2, 1.5 mol of HCl and 0.10 mol of CC14. The following reaction becomes possible:
Cl₂(g) + CHCl3 (g) →HC1 (g) +CC14 (8)
The equilibrium constant K for this reaction is 0.642 at the temperature of the flask.
Calculate the equilibrium molarity of CHC13. Round your answer to two decimal places.
M
X
Ś

Expert Solution

Step 1: Determine initial concentration of all species

Volume of the flask = 250ml

$space equals space space 250 over 1000 l i t space equals space 0.25 space l i t$

Hence initial concentration of Cl₂

$left square bracket C l subscript 2 right square bracket equals space fraction numerator m o l over denominator V o l u m e space i n space L end fraction equals space fraction numerator 1.8 space m o l over denominator 0.25 L end fraction space equals space 6.4 space M left square bracket H C l right square bracket space equals space fraction numerator 1.5 m o l over denominator 0.25 L end fraction space equals space 6 space M space space left square bracket C H C l subscript 3 right square bracket space equals fraction numerator 0 m o l over denominator 0.25 L end fraction space equals space 0 space space a n d space left square bracket C C l subscript 4 right square bracket space equals space fraction numerator 0.1 m o l over denominator 0.25 L end fraction space equals space 0.4 space M space$

since from the reaction we can see that if 1 mole of HCl is formed than 1 mole of CCl₄ will be formed and 1 moles of CHCl₃ will be reacting with 1 mole of Cl₂

Since volume is constant hence we can say same thing about the concentration as well

i.e if 1 M concentration of HCl is formed than 1 M concentration of CCl₄ will be formed and 1 M concentration of CHCl₃ will be reacting with 1 M concentration of Cl₂

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