/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Predict the shift in the equilib... [FREE SOLUTION] | 91Ó°ÊÓ

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Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
The equilibrium shifts for each reaction when the volume of the reaction container is increased are as follows: a. Shifts to the left b. Shifts to the right c. No shift d. Shifts to the right e. Shifts to the right

Step by step solution

01

Analyze the change in moles of gas

First, let's analyze the change in moles of gas on both sides for each reaction: a. 1 mole + 3 moles \(\to\) 2 moles b. 1 mole \(\to\) 1 mole + 1 mole c. 1 mole + 1 mole \(\to\) 2 moles d. 1 mole \(\to\) 1 mole + 1 mole e. 0 moles \(\to\) 0 moles + 1 mole
02

Determine the direction of the equilibrium shift

Let's determine the direction of the equilibrium shift for each reaction in response to the increase in volume (decrease in pressure): a. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the reactants (left) side. b. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the products (right) side. c. Since the number of moles of gas is the same on either side, increasing the volume will not cause a shift in the equilibrium position. d. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the products (right) side. e. Increasing the volume causes the system to shift towards the side with more moles of gas, which in this case is the products (right) side. So, the shifts in the equilibrium positions for each reaction are as follows: a. Shifts to the left b. Shifts to the right c. No shift d. Shifts to the right e. Shifts to the right

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Most popular questions from this chapter

At a particular temperature, a 3.0-L flask contains \(2.4 \mathrm{~mol} \mathrm{Cl}_{2}\), \(1.0 \mathrm{~mol} \mathrm{NOCl}\), and \(4.5 \times 10^{-3} \mathrm{~mol}\) NO. Calculate \(K\) at this temperature for the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

At \(1100 \mathrm{~K}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ What is the value of \(K\) at this temperature?

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ The initial density of the system was recorded as \(4.495 \mathrm{~g} / \mathrm{L}\). After equilibrium was reached, the density was noted to be \(4.086 \mathrm{~g} / \mathrm{L}\). a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

In a study of the reaction $$ 3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) $$ at \(1200 \mathrm{~K}\) it was observed that when the equilibrium partial pressure of water vapor is \(15.0\) torr, that total pressure at equilibrium is \(36.3\) torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at \(1200 \mathrm{~K}\). (Hint: Apply Dalton's law of partial pressures.)

Consider the reaction $$ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) $$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{~K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{~K}\), the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of \(1.00 \mathrm{~atm} .\) Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

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