Question #47dc1

Start by writing the balanced chemical equation that describes this equilibrium reaction

#"H"_ (2(g)) + "Cl"_ (2(g)) rightleftharpoons color(red)(2)"HCl"_((g))#

Here hydrogen gas, #"H"_2#, reacts with chlorine gas, #"Cl"_2#, to form hydrogen chloride, #"HCl"#.

Notice that the reaction produces #color(red)(2)# moles of hydrogen chloride for every #1# mole of hydrogen gas and #1# mole of chlorine gas that take part in the reaction.

Now, the equilibrium constant for this reaction, #K_c#, is defined as the ratio that exists between the equilibrium concentration of the product and the equilibrium concentrations of the two reactants, all raised to the power of their respective stoichiometric coefficients.

#K_c = (["HCl"]^color(red)(2))/(["H"_2] * ["Cl"_2])#

In your case, the equilibrium constant at the temperature at which the reaction takes place is equal to #2.5 * 10^(34)#.

Right from the start, the magnitude of the equilibrium constant tells you that the equilibrium will lie almost entirely to the right, i.e. the reaction will almost go to completion.

The forward reaction, i.e. the reaction that produces hydrogen chloride, will be favored. This means that at equilibrium, you can expect the concentration of hydrogen chloride to be significantly higher than the concentrations of the two reactants.

Rearrange the expression of #K_c# to isolate #["HCl"]# on one side of the equation

#["HCl"]^color(red)(2) = K_c * ["H"_2] * ["Cl"_2]#

This will give you

#["HCl"] = color(red)(sqrt(color(black)(K_c * ["H"_2] * ["Cl"_2]))#

Plug in your values to get

#["HCl"] = sqrt(2.5 * 10^(34) * 3.8 * 10^(-5) * 4.6 * 10^(-6))#

#["HCl"] = color(green)(|bar(ul(color(white)(a/a)2.1 * 10^(12)"M"color(white)(a/a)|)))#

The answer is rounded to two sig figs.

As predicted, the equilibrium concentration of hydrogen chloride is indeed significantly higher than the equilibrium concentrations of the two reactants.