The equilibrium constant \(K_{\mathrm{c}}\) for the reaction \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is 0.83 at \(375^{\circ} \mathrm{C} . \mathrm{A}\) 14.6-g sample of ammonia is placed in a \(4.00-\mathrm{L}\) flask and heated to \(375^{\circ} \mathrm{C}\). Calculate the concentrations of all the gases when equilibrium is reached.

Short Answer

Expert verified
The solution involves setting up an ICE table, substituting the concentrations into the equilibrium expression and solving the resulting quadratic equation. Perform the appropriate calculations to find the equilibrium concentrations of NH3, N2, and H2.

Step by step solution

01

Interpret the Problem and Gather Data

The reaction is \(2 NH_{3}(g) \rightleftharpoons N_{2}(g) + 3 H_{2}(g)\) with equilibrium constant \(K_{c} = 0.83\). Start with a 14.6g sample of ammonia in a 4.00L. flask. To convert grams of NH₃ to moles, remember \(NH_{3}\)'s molar mass is about 17 g/mol.
02

Set up ICE Table

An ICE table helps organize the change in concentrations throughout the reaction. In the table, I initial), C (change), and E (equilibrium) refer to the concentrations at these instances.\n\nInitial moles of ammonia = 14.6g / 17 g/mol = 0.859 moles. Thus, initial concentrations: [NH3] = 0.859 mol/4.00 L = 0.215 M, [N2] = [H2] = 0 M (none initially).\n\nThe changes in concentration will be: -2x for NH3 (as per the stoichiometric coefficient), +x for N2, and +3x for H2.\n\nThe equilibrium concentrations thus become: [NH3] = 0.215 - 2x M, [N2] = x M, and [H2] = 3x M. Here x is the change in concentration of N2 or 1/3 of the change in concentration of H3.
03

Substitute the Values in the Expression for Kc

The expression for \(K_{c}\) is [N2][H2]^3 / [NH3]^2. Substitute the equilibrium concentrations into Kc expression and set it equal to 0.83: 0.83 = (x(3x)^3) / ((0.215-2x)^2). This simplifies into a quadratic equation.
04

Solve the Quadratic Equation

Solving the quadratic equation will provide the value of x. Then substitute x back into the equilibrium concentrations to find the final values.\n\nSolving the quadratic equation will involve either factoring, completing the square, or using the quadratic formula.
05

Check Your Answer

After solving, it's important to verify the answer by substituting the calculated values back into the \(K_{c}\) equation to verify that it gives the known \(K_{c}\) value of 0.83.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g)$$ at the same temperature.

Consider the reaction $$\begin{array}{r}2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \\\\\Delta H^{\circ}=-198.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to (a) increase the temperature; (b) increase the pressure; (c) increase \(\mathrm{SO}_{2}\); (d) add a catalyst; (e) add helium at constant volume.

Consider the gas-phase reaction $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$$ Predict the shift in the equilibrium position when helium gas is added to the equilibrium mixture at (a) constant pressure and (b) constant volume.

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant \(K_{P}\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment \(2.00 \mathrm{~mol} \mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol} \mathrm{O}_{2}\) were initially present in a flask. What must the total pressure at equilibrium be in order to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Define reaction quotient. How does it differ from equilibrium constant?

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