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Chapter 15: Problem 18

A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M},\left[\mathrm{N}_{2}\right]=0.11 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=1.91 \mathrm{M}\). Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)

Short Answer

Expert verified
For the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\), the equilibrium constant \(K_{c}\) is calculated to be a certain value. When the stoichiometric coefficients are halved, the \(K_{c}\) for the new reaction (\(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)) was computed to be the square root of the original equilibrium constant.

Step by step solution

01

Solve for Kc Using the Provided Concentrations and Reaction for Part (a)

For a general reaction of the form \(Aa + Bb \rightleftharpoons Cc + Dd\), the equilibrium constant \(K_{c}\) is given by \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\). In this case, for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2\mathrm{NH}_{3}(g)\), \(K_{c}\) will be equal to \(\frac{[\mathrm{NH}_{3}]^{2}}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}\). We substitute the given concentrations into this formula to get \(K_{c}=\frac{(0.25)^{2}}{(0.11)(1.91)^{3}}\) and solve this to get the value of \(K_{c}\).
02

Solve for Kc using the Provided Concentrations and Reaction for Part (b)

For part (b) of the exercise, we have different stoichiometric coefficients. Since the form is now \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\), the formula for \(K_{c}\) will be \(\frac{[\mathrm{NH}_{3}]}{[\mathrm{N}_{2}]^{1/2}[\mathrm{H}_{2}]^{3/2}}\). We substitute the given concentrations into this expression to get \(K_{c}=\frac{(0.25)}{(0.11)^{1/2}(1.91)^{3/2}}\) and solve this to get the value of \(K_{c}\).
03

Reflect on the Effect of Coefficients on the Equilibrium Constant

Compare the results obtained in Steps 1 and 2, and note how modifying the stoichiometric coefficients of a balanced equation affects the equilibrium constant. It's interesting to note that when we halve the coefficients in the equation, the equilibrium constant for the new equation becomes the square root of the original. This illustrates an important property of equilibrium constants: if the balanced equation for a reaction is multiplied by a factor, the equilibrium constant for the new equation is the original constant, raised to the power of that factor.

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Most popular questions from this chapter

Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C}\), the total pressure of the system is found to be 4.50 atm. If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\)

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

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right].\)

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00 -L flask. At \(648 \mathrm{~K}\), there is 0.0345 mole of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$
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