Question

In: Chemistry

A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL...

A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.

A (aq) + 2B (aq) <---> C (aq)

Solutions

Expert Solution

Reactions which go for completion are called irreversible reaction. While there are few reactions which do not go for completion and hence, known as reversible reaction.

In these reversible reactions, the reactants form products in a forward reaction and the products formed can get change back to reactants in the backward reaction.

A State known as equilibrium state is reached wherein reactants and products are present but their concentrations do not seem to change with time. This state is known as state of chemical equilibrium.

For a chemical reaction, chemical equilibrium is achieved when rate of forward reaction becomes equals to rate of background reaction. In this state concentrations of both reactants and products do not change with time if reaction conditions remain same.

For a given elementary reaction,

Rate of forward reaction = k1[A]a[B]b...............(1)

Rate of backward reaction = k2[C]c[D]d...............(2)

Where,

k1 = Rate constant of forward reaction

k2 = Rate constant of backward reaction

On dividing equation (1) by (2), we get

At equilibrium,

Rate of forward reaction = Rate of background reaction

Where K is equilibrium constant

In the given reaction, one mole of A reacts with 2 moles of B to form C.

For the above reaction, expression for equilibrium constant is:

Given,

Amount of A = 30.0 mL of 0.150 M

Amount of B = 25.0 mL of 0.200 M

Equilibrium concentration of C = 0.045 M

Volume of the solution = 30.0 mL + 25.0 mL = 55 mL

Given equilibrium concentration of C is 0.0454M

Concentration of A at equilibrium = (0.08181 – 0.0454)M = 0.0364 M

Concentration of B at equilibrium = (0.09090 – 2*0.0454)M

=(0.09090 – 0.0908)M = 0.0001 M

Using the expression,

Equilibrium constant for the given reaction is 1.25 *108.


Related Solutions

A solution is made by mixing 25.0 mL of 0.250 M nitric acid (HNO3) with 15.0...
A solution is made by mixing 25.0 mL of 0.250 M nitric acid (HNO3) with 15.0 mL of 0.500 M sodium hydroxide (NaOH). What is the resulting concentration of each ion in solution?
A solution is made by mixing 200.0 ml of 0.300 M FeI3 (436.56 g / mol)...
A solution is made by mixing 200.0 ml of 0.300 M FeI3 (436.56 g / mol) with 130.0 ml of 0.800 M LiBr (86.8450 g/mol) with 150.0 ml of a 0.250M MgI2 (293.89 g/ mol) . What is the molar concentration of iodide ion in the new solution?
Calculate the pH of a solution prepared by mixing 20 ml of 9.12 mol/L salicylic acid...
Calculate the pH of a solution prepared by mixing 20 ml of 9.12 mol/L salicylic acid with 20 ml water
if 30.0 ml of 0.150M CaCL2 is added to 15 ml of 0.100M AgNO3. what is...
if 30.0 ml of 0.150M CaCL2 is added to 15 ml of 0.100M AgNO3. what is the mass in grams of AgCL precipitate?
50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both...
50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both at 20.0 degree Celsius, were mixed in a calorimeter. Calculate the molar heat of neutralization of HNO3 in kJ/mol if: (1) final temperature was 28.9 degree Celsius; (2) the mass of the overall solution was 102.0 g; (3) the heat capacity of the calorimeter was 25.0 J/C; (4) and assume that the specific heat of solution is the same as water, 4.184 J/(g C);
What is the pH of the solution made by mixing 0.2 mol NaH2PO4 and 0.5 mol...
What is the pH of the solution made by mixing 0.2 mol NaH2PO4 and 0.5 mol NaOH with water to make 1.00 L of solution? The pKa values for H3PO4 are 2.12 7.20 and 12.
In the titration of 30.0 mL of a 0.200 M solution of a hypothetical compound NaH2M,...
In the titration of 30.0 mL of a 0.200 M solution of a hypothetical compound NaH2M, what is the pH of the solution after the addition of 30.0 mL of 0.100 M NaOH? For H3M, pKa1 = 3.00, pKa2 = 6.00, and pKa3 = 9.00.? The answer is 6.0 which is not what I got in my work. Please help me out !
Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1000...
Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.1000 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL 0.100 M Ba(OH)2, and 10.0 mL of 0.150 M KOH. Calculte the pH of this solution. Ka (HOCl) = 3.5 x 10-8 What is the pH?
Calculate the pH of a solution prepared by mixing a) 25.0 mL of 0.512 M NaOH...
Calculate the pH of a solution prepared by mixing a) 25.0 mL of 0.512 M NaOH and 34.0 mL of 0.187 M HCl b) 46.0 mL of 0.235 M KOH and 50.0 mL of 0.420 M HC3H5O2 c) 400 mL of 0.250 M NH3 and 250 mL of 0.120 M HCl
Calculate the pH of a solution prepared by mixing a) 25.0 mL of 0.512 M NaOH...
Calculate the pH of a solution prepared by mixing a) 25.0 mL of 0.512 M NaOH and 34.0 mL of 0.187 M HCl b) 46.0 mL of 0.235 M KOH and 50.0 mL of 0.420 M HC3H5O2 c) 400 mL of 0.250 M NH3 and 250 mL of 0.120 M HCl
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT