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Chapter 15: Problem 101

In the reversible reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{HI}(\mathrm{g}),\) an initial mixture contains \(2 \mathrm{mol} \mathrm{H}_{2}\) and 1 mol I \(_{2} .\) The amount of HI expected at equilibrium is (a) \(1 \mathrm{mol} ;\) (b) \(2 \mathrm{mol} ;\) (c) less than \(2 \mathrm{mol}\); (d) more than 2 mol but less than 4 mol.

Short Answer

Expert verified
The correct option explaining the amount of HI expected at equilibrium is option (b), which states it would be 2 mol.

Step by step solution

01

Understand the Reversible Reaction

First, comprehend that the reaction is reversible indicating that both the forward (reactants forming products) and backward (products forming reactants) reactions are occurring concurrently when the system is at equilibrium. The equation for this reaction is: \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\). The reactants are hydrogen and iodine, which form hydrogen iodide in the product.
02

Determine the Initial Amount of Each Species

As per the exercise, the initial amounts are 2 mol of hydrogen (\(H_{2}\)) and 1 mol of iodine (\(I_{2}\)). At the start, no hydrogen iodide is present, hence its initial amount is 0 mol.
03

Evaluate the Stoichiometry and Equilibrium

From the balanced equation, for every one mol of \(H_2\) and \(I_2\) that react, they form 2 mol of \(HI\). However, initially there is an excess of hydrogen which means that all 1 mol of \(I_2\) will be consumed, producing 2 mol of \(HI\) in the process. After this, no more \(HI\) can be produced as \(I_2\) is now the limiting reactant.
04

Select the right answer

From the previous steps, it is established that only 2 mol of \(HI\) can be produced. This aligns with option (b).

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Most popular questions from this chapter

Starting with \(0.280 \mathrm{mol} \mathrm{SbCl}_{3}\) and \(0.160 \mathrm{mol} \mathrm{Cl}_{2},\) how many moles of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present when equilibrium is established at \(248^{\circ} \mathrm{C}\) in a 2.50 L flask? $$\begin{aligned} \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 2.5 \times 10^{-2} \mathrm{at} \ 248^{\circ} \mathrm{C} \end{aligned}$$

A mixture consisting of \(0.150 \mathrm{mol} \mathrm{H}_{2}\) and \(0.150 \mathrm{mol} \mathrm{I}_{2}\) is brought to equilibrium at \(445^{\circ} \mathrm{C},\) in a 3.25 L flask. What are the equilibrium amounts of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and HI? $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\ \mathrm\ {at}\ 445^{\circ} \mathrm{C}$$

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Châtelier's principle; (d) effect of a catalyst on equilibrium.

In the equilibrium described in Example \(15-12,\) the percent dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be expressed as $$\frac{3.00 \times 10^{-3} \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}}{0.0240 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4} \text { initially }} \times 100 \%=12.5 \%$$ What must be the total pressure of the gaseous mixture if \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) is to be \(10.0 \%\) dissociated at \(298 \mathrm{K} ?\) $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.113 \text { at } 298 \mathrm{K} $$

Given the equilibrium constant values $$\begin{aligned} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{c}}=2.7 \times 10^{-18} \\ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) & \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) K_{\mathrm{c}}=4.6 \times 10^{-3} \\ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.1 \times 10^{-9} \end{aligned}$$ Determine a value of \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$
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