Video Transcript
Ammonia can be produced by the reaction between hydrogen and nitrogen gas that is
represented by the following equation. Three H2 plus N2 is in equilibrium with two NH3. 5.00 moles of nitrogen gas and 5.00 moles of hydrogen gas are placed in a sealed
vessel that has a volume of 20.00 cubic decimeters at 500 kelvin. At equilibrium, only 0.25 moles of nitrogen would be converted into ammonia. Calculate the value of 𝐾 𝑐 at this temperature, giving the answer to three decimal
places.
𝐾 𝑐 is the equilibrium constant for concentration. The equilibrium constant for concentration expresses the relationship between the
reactant and product concentrations at equilibrium. Let’s consider a generic reaction equation where the lowercase letters represent
stoichiometric coefficients and the uppercase letters represent chemical
formulas. Using the generic reaction equation, we can write an expression for the equilibrium
constant for concentration. In this expression, the concentrations of the products at equilibrium are raised to
their stoichiometric coefficients, then divided by the concentrations of the
reactants at equilibrium raised to their stoichiometric coefficients.
We can use the generic equations to help us write a formula that would allow us to
calculate the 𝐾 𝑐 for the given reaction. This gives us the equation 𝐾 𝑐 equals the concentration of NH3 squared divided by
the concentration of H2 cubed times the concentration of N2. We now have an equation to calculate the equilibrium constant for concentration for
this reaction. But the question doesn’t provide us with the equilibrium concentrations of the
reactants or products, so we’ll need to calculate them. To help us, we can use the ICE method. In this method, we’ll record the initial number of moles of each species involved in
the reaction, the change in the number of moles, and the number of moles at
equilibrium.
We are told in the question that 5.00 moles of nitrogen gas and 5.00 moles of
hydrogen gas are placed in a sealed vessel. Initially, there will be zero moles of NH3, the product. We can denote the change in the number of moles as 𝑥. The amount of each reactant will decrease and the amount of the product will
increase. The change in the number of moles should reflect the stoichiometric coefficients in
the chemical equation. So, the amount of hydrogen will decrease by three 𝑥, while the amount of ammonia
will increase by two 𝑥. We are told in the question that only 0.25 moles of nitrogen would be converted into
ammonia. As the stoichiometric coefficient of nitrogen is one, its change in amount is minus
𝑥. Therefore, 0.25 moles must be the value of 𝑥.
We can then substitute 0.25 for 𝑥 and calculate the change in amount for each
species. So, the amount of hydrogen will decrease by 0.75 moles. The amount of nitrogen will decrease by 0.25 moles. And the amount of ammonia will increase by 0.5 moles. If we add the change in amount to the initial amount, we can calculate the
equilibrium amount. We find that at equilibrium there are 4.25 moles of hydrogen, 4.75 moles of nitrogen,
and 0.5 moles of ammonia. But, to use the 𝐾 𝑐 equation, we need to know the concentration of each species,
not the amount. We are told in the question that the reaction takes place in a sealed vessel with a
volume of 20.00 cubic decimeters. So to determine the equilibrium concentrations, we simply need to divide each
equilibrium amount by 20.00 cubic decimeters.
Performing this calculation gives us the equilibrium concentration of each
species. Now, we can calculate the equilibrium constant for concentration. We can substitute the equilibrium concentrations into the equation for 𝐾 𝑐. We then need to square the concentration of ammonia and cube the concentration of
hydrogen. Multiplying the values in the denominator gives us 2.2789 times 10 to the negative
third moles to the fourth per decimeter to the 12th. Performing the final step of the calculation gives us a numerical answer of
0.27425.
But what about the unit? The unsimplified or complex unit is moles squared per decimeter to the sixth divided
by moles to the fourth per decimeter to the 12th. We can simplify this complex fraction by multiplying both the numerator and
denominator by decimeters to the 12th per moles to the fourth. This will cancel the numerator of the complex fraction. This action will also allow some of the mole and decimeter units to cancel. This leaves us with the unit decimeter to the sixth per mole squared. This unit can be rewritten in a single line as moles to the negative second times
decimeters to the sixth.
Finally, to complete the problem, we need to give our answer to three decimal
places. Rounding appropriately, we have determined that the value of the equilibrium constant
for concentration for the reaction between hydrogen and nitrogen gas is 0.274 moles
to the negative second times decimeters to the sixth.
