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Chapter 14: Problem 15

(a) Rewrite this reaction with the word heat in it: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{CO}_{2}(g) \quad \Delta E=-563.5 \mathrm{~kJ} $$ (b) Which way does the reaction shift when the temperature is raised? Explain your answer.

Short Answer

Expert verified
(a) The rewritten reaction including "heat" is: \[ 2 CO(g) + O_{2}(g) \rightleftarrows 2 CO_{2}(g) + \text{heat} \] (b) When the temperature is raised, the reaction will shift to the left (towards CO and O2) to absorb the added heat, according to Le Chatelier's principle.

Step by step solution

01

I. Identify the heat term in the reaction

The reaction has a ΔE value of -563.5 kJ, which represents the energy released by the reaction as heat.
02

II. Rewrite the reaction including the term "heat"

Since ΔE is negative, the reaction releases energy as heat. We can include this in the reaction as follows: \[ 2 CO(g) + O_{2}(g) \rightleftarrows 2 CO_{2}(g) + \text{heat} \] #b. Identify the direction in which the reaction will shift when the temperature is raised#
03

I. Understand the role of energy in an equilibrium reaction

When energy is added to an equilibrium reaction, the system adjusts itself to minimize the change, in accordance with Le Chatelier's principle. In this case, the added heat will cause the reaction to shift in the direction that absorbs the extra heat.
04

II. Determine which direction the reaction shifts

Since the reaction releases heat when it proceeds to form CO2, the opposite direction (formation of CO and O2) will absorb heat. Therefore, when the temperature is raised, the reaction will shift to the left (towards the reactants CO and O2) to absorb the added heat.
05

III. Explain the answer

In summary, when the temperature is raised, the equilibrium reaction: \[ 2 CO(g) + O_{2}(g) \rightleftarrows 2 CO_{2}(g) + \text{heat} \] will shift to the left, favoring the formation of CO and O2, in order to absorb the added heat and minimize the change in temperature, according to Le Chatelier's principle.

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Most popular questions from this chapter

In which direction does the reaction \(\mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) \(\rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q)\) shift when: (a) \(\left[\mathrm{CO}_{2}\right]\) is increased? (b) The volume of the reaction vessel is decreased? (c) \(\mathrm{Ca}^{2+}(a q)\) is added? (d) \(\mathrm{CaCO}_{3}(s)\) is removed?

Write the equilibrium constant expression for: (a) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(g)\) \(\rightleftarrows \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)\) \(\rightleftarrows \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{I}_{2}(s) \rightleftarrows \mathrm{I}_{2}(g)\) (e) \(\mathrm{TiCl}_{4}(g)+2 \mathrm{Mg}(s) \rightleftarrows \mathrm{Ti}(s)+2 \mathrm{MgCl}_{2}(s)\) (f) \(\mathrm{Ni}(\mathrm{OH})_{2}(s) \rightleftarrows \mathrm{Ni}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\)

What is a heterogeneous chemical reaction? Where does a heterogeneous reaction occur?

The reaction \(\mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) is run in a 10.0-L vessel. The vessel is loaded with 1 mole of \(\mathrm{CO}\) and 3 moles of \(\mathrm{H}_{2}\). At equilibrium, the amounts are \(0.613\) mole of \(\mathrm{CO}, 1.839\) moles of \(\mathrm{H}_{2}, 0.387\) mole of \(\mathrm{CH}_{4}\), and \(0.387\) mole of \(\mathrm{H}_{2} \mathrm{O}\). What is the value of the equilibrium constant for this reaction? Describe the position of the equilibrium.

The water in a beaker of water left in a room will slowly evaporate until the beaker is dry. However, place that same beaker in a sealed box and the water level in the beaker will drop a bit but then remain constant. Is the latter case an example of equilibrium? Explain your answer.
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