Question Video: Calculating the Equilibrium Constant for Concentration Given the Initial Amount of Each Reactant | Nagwa Question Video: Calculating the Equilibrium Constant for Concentration Given the Initial Amount of Each Reactant | Nagwa

Question Video: Calculating the Equilibrium Constant for Concentration Given the Initial Amount of Each Reactant Chemistry • Third Year of Secondary School

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Ammonia can be produced by the reaction between hydrogen and nitrogen gas that is represented by the following equation: 3 H₂ + N₂ ⇌ 2 NH₃. 5.00 mol of nitrogen gas and 5.00 mol of hydrogen gas are placed in a sealed vessel that has a volume of 20.00 dm³ at 500 K. At equilibrium, only 0.25 mol of nitrogen would be converted into ammonia. Calculate the value of 𝐾_𝑐 at this temperature, giving the answer to 3 decimal places.

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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.

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