Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [N2]eq = 1.5 M, [H2]eq = 1.1 M, [NH3]eq = 0.47 M. N2(g) + 3 H2(g) 2 NH3(g)

Answer : The value of equilibrium constant (K) is, 0.11

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given chemical reaction is:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

Thus, the value of equilibrium constant (K) is, 0.11

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.