In a chemical reaction, substrate molecule A is broken down to form one molecule of product B and one molecule of product C. The equilibrium constant, K, for this reaction is 0.5. If we start with a mixture containing only substrate A at a concentration of 1 M, what will be the concentration of A when the reaction reaches equilibrium?a. 0.125 Mb. 0.25 Mc. 0.333 Md. 0.5 Me. 0.667 M

The answer to your question is d. 0.5 M

Explanation:

Data

[A] = 1M

K = 0.5

Concentration of B and C at equilibrium = x

Concentration of A at equilibrium = 1 - x

Equation of equilibrium

                                       k =

Substitution

Simplification

                                       0.5 =

Solve for x

                                      0.5(1 - x) = x²

                                      0.5 - 0.5x = x²

                                       x² + 0.5x - 0.5 = 0

Find the roots             x₁ = 0.5     x₂ = -1

There are no negative concentrations so the concentration of A at equilibrium is  

                       [A] = 1 - 0.5

                             = 0.5 M

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :