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Chapter 15: Problem 32

In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?

Short Answer

Expert verified
a) No, the mixture cannot be at equilibrium because the calculated equilibrium constant does not equal the given one. They will only equal when the reaction is at equilibrium. b) A net change will occur in the direction of forming more CO2 and H2.

Step by step solution

01

Understanding the Problem

In this exercise, a chemical reaction is given along with its equilibrium constant at a certain temperature. Equal masses of each reactant and product are taken in a reaction vessel at those same conditions: CO(g) + H2O(g) ↔ CO2(g) + H2(g), and K = 31.4 at 588K. It is then required to find out whether or not the mixture is in equilibrium, and if it is not, in what direction the reaction will proceed.
02

Determining the Initial State

Since it is mentioned that equal masses of each reactant and product are present initially, we can assume that their concentrations are also equal as chemical concentrations depend on masses and not on the nature of the substances. Let's assign this common concentration as 'a'.
03

Determining the Equilibrium Constant

Next, we can write our equilibrium constant expression as per the law of mass action. For the reaction CO(g) + H2O(g) ↔ CO2(g) + H2(g), the equilibrium constant K= [CO2][H2] / [CO][H2O]. As per the given condition, we have K = (a^2) / (a^2).
04

Comparing Given and Calculated Equilibrium Constants

The calculated equilibrium constant should equal the value that is given, which is 31.4. However, our calculated K is equal to 1, because (a^2) / (a^2) simplifies directly to 1.
05

Conclusion

Based on the comparison, we see that the given chemical system is not in equilibrium, because the calculated K does not match the given K. Therefore, the reaction needs to proceed in the direction that makes the two equal.
06

Direction of Reaction

Since calculated K (which equals 1) is less than the given K (which is 31.4), the reaction needs to proceed in the direction that increases the value of K. This is the direction that forms more products, which is the forward direction. So, the answer to (b) is that a net change will occur in the direction of forming more CO2 and H2.

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Most popular questions from this chapter

An equilibrium mixture at 1000 K contains an equilibrium mixter \(0.276\ \mathrm{mol}\ \mathrm{H}_{2}, 0.276 \mathrm{mol}\ \mathrm{CO}_{2}, 0.224\ \mathrm{mol}\ \mathrm{CO},\) and \(0.224\ \mathrm{mol}\ \mathrm{H}_{2} \mathrm{O}\) $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Show that for this reaction, \(K_{\mathrm{c}}\) is independent of the reaction volume, \(V\) (b) Determine the value of \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\)

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

The decomposition of \(\mathrm{HI}(\mathrm{g})\) is represented by the equation $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$$ \(\mathrm{HI}(\mathrm{g})\) is introduced into five identical \(400 \mathrm{cm}^{3}\) glass bulbs, and the five bulbs are maintained at \(623 \mathrm{K}\) Each bulb is opened after a period of time and analyzed for \(I_{2}\) by titration with \(0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) $$\begin{array}{l} \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \end{array}$$ Data for this experiment are provided in the table below. What is the value of \(K_{\mathrm{c}}\) at \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

The following reaction is an important reaction in the citric acid cycle: citrate(aq) \(+\mathrm{NAD}_{\mathrm{ox}}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons\) \(\mathrm{CO}_{2}(\mathrm{aq})+\mathrm{NAD}_{\mathrm{red}}+\) oxoglutarate \((\mathrm{aq}) \quad K=0.387\) Write the equilibrium constant expression for the above reaction. Given the following data for this reaction, \([\text { citrate }]=0.00128 \mathrm{M},\left[\mathrm{NAD}_{\mathrm{ox}}\right]=0.00868,\left[\mathrm{H}_{2} \mathrm{O}\right]=\) \(55.5 \mathrm{M},\left[\mathrm{CO}_{2}\right]=0.00868 \mathrm{M},\left[\mathrm{NAD}_{\mathrm{red}}\right]=0.00132 \mathrm{M}\) and [oxoglutarate] \(=0.00868 \mathrm{M},\) calculate the reaction quotient. Is this reaction at equilibrium? If not, in which direction will it proceed?
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