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Chapter 13: Problem 23

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) d. \(2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)

Short Answer

Expert verified
a. For the first reaction, the equilibrium constant is \( K_{new} = (1.3 \times 10^{-2})^{(1/2)}\). b. For the second reaction, the equilibrium constant is \( K_{new} = \frac{1}{1.3 \times 10^{-2}}\). c. For the third reaction, the equilibrium constant is \( K_{new} = K_{temp}^{(1/2)}\), where \(K_{temp} = \frac{1}{1.3 \times 10^{-2}}\). d. For the fourth reaction, the equilibrium constant is \( K_{new} = (1.3 \times 10^{-2})^{(2)}\).

Step by step solution

01

Identify the factor by which the coefficients have changed

In this case, all the stoichiometric coefficients have been divided by 2. Hence, the factor is 1/2.
02

Calculate the new equilibrium constant

We take the given equilibrium constant, K = 1.3 x 10^{-2}, and raise it to the power of the factor (1/2) we found in Step 1: \(K_{new} = K^{(1/2)} = (1.3 \times 10^{-2})^{(1/2)}\) b. Second reaction: \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\)
03

Identify the order of the reaction

This reaction has the reverse order of the given reaction.
04

Calculate the new equilibrium constant

To find the equilibrium constant for the reverse reaction, we take the reciprocal of the given equilibrium constant: \(K_{new} = \frac{1}{K} = \frac{1}{1.3 \times 10^{-2}}\) c. Third reaction: \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g) +\frac{3}{2} \mathrm{H}_{2}(g)\)
05

Factor between coefficients

All coefficients have been divided by 2, so the factor is 1/2 again.
06

Reverse the reaction

The reaction is reversed compared to the given one, so we take the reciprocal first: \(K_{temp} = \frac{1}{K} = \frac{1}{1.3 \times 10^{-2}}\)
07

Calculate K for the reversed reaction with factor 1/2

We take K_temp and raise it to the power of the factor, (1/2): \(K_{new} = K_{temp}^{(1/2)}\) d. Fourth reaction: \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)
08

Identify the factor by which the coefficients have changed

In this case, all the stoichiometric coefficients have been multiplied by 2. Hence, the factor is 2.
09

Calculate the new equilibrium constant

We take the given equilibrium constant, K = 1.3 x 10^{-2}, and raise it to the power of the factor (2) we found in Step 1: \(K_{new} = K^{(2)} = (1.3 \times 10^{-2})^{(2)}\)

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Most popular questions from this chapter

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3}\), what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

For the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ at \(600 . \mathrm{K}\), the equilibrium constant, \(K_{\mathrm{p}}\), is \(11.5 .\) Suppose that \(2.450 \mathrm{~g} \mathrm{PCl}_{5}\) is placed in an evacuated \(500 .-\mathrm{mL}\) bulb, which is then heated to \(600 . \mathrm{K}\). a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of \(\mathrm{PCl}_{5}\) at equilibrium?

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \quad K=2.2\) \(\begin{array}{ll}\text { Acetic acid } \text { Ethanol } & \text { Ethyl acetate }\end{array}\) For the following mixtures \((a-d)\), will the concentration of \(\mathrm{H}_{2} \mathrm{O}\) increase, decrease, or remain the same as equilibrium is established? a. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.010 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.010 \mathrm{M}\) b. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.0020 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.0020 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.10 \mathrm{M}\) c. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.88 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.12 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.044 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=6.0 \mathrm{M}\) d. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=4.4 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=4.4 M\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.88 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=10.0 \mathrm{M}\) e. What must the concentration of water be for a mixture with \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=2.0 \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.10 M\), and \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=5.0 M\) to be at equilibrium? f. Why is water included in the equilibrium expression for this reaction?

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{~K}\) at an initial pressure of \(1.00 \mathrm{~atm}\), where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$ \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) $$ At equilibrium, the partial pressure of \(\mathrm{S}_{\mathrm{g}}\) is \(0.25 \mathrm{~atm} .\) Calculate \(K\). for this reaction at \(1325 \mathrm{~K}\).

Peptide decomposition is one of the key processes of digestion, where a peptide bond is broken into an acid group and an amine group. We can describe this reaction as follows: Peptide \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) acid \(\operatorname{group}(a q)+\) amine \(\operatorname{group}(a q)\) If we place \(1.0\) mole of peptide into \(1.0 \mathrm{~L}\) water, what will be the equilibrium concentrations of all species in this reaction? Assume the \(K\) value for this reaction is \(3.1 \times 10^{-5}\).
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