/*! 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 97 (a) For each of the following re... [FREE SOLUTION] | 91影视

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(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1 .\) (c) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. $$ \begin{array}{l}{\text { (i) } 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)} \\ {\text { (ii) } 2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)} \\ {\text { (iii) } \mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)} \\ {\text { (iv) } 2 \mathrm{V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{V}(s)+5 \mathrm{O}_{2}(g)}\end{array} $$

Short Answer

Expert verified
The key findings of the analysis are: (i) Mg(s) + O鈧(g) 鈫 2MgO(s): 螖H鈦 < 0, 螖S鈦 < 0, K > 1 at low temperatures, K decreases as temperature increases. (ii) 2KI(s) 鈫 2K(g) + I鈧(g): 螖H鈦 > 0, 螖S鈦 > 0, K > 1 at high temperatures, K increases as temperature increases. (iii) Na鈧(g) 鈫 2Na(g): 螖H鈦 > 0, 螖S鈦 > 0, K > 1 at high temperatures, K increases as temperature increases. (iv) 2V鈧侽鈧(s) 鈫 4V(s) + 5O鈧(g): 螖H鈦 > 0, 螖S鈦 > 0, K > 1 at high temperatures, K increases as temperature increases.

Step by step solution

01

(i) Reaction 1: Predicting 螖H鈦 and 螖S鈦

Mg(s) reacts with O鈧(g) to form MgO(s). Since the product MgO is more stable than the reactants, the reaction is likely exothermic, meaning that 螖H鈦 will be negative. The reaction also decreases the number of gas molecules, implying a decrease in entropy (螖S鈦 < 0).
02

(i) Reaction 1: Predicting K > 1 and the Temperature Dependence of K

Since both 螖H鈦 and 螖S鈦 are negative for this reaction, its spontaneity is favored at lower temperatures (K > 1), and K decreases as the temperature increases.
03

(ii) Reaction 2: Predicting 螖H鈦 and 螖S鈦

In this reaction, solid potassium iodide dissociates into gaseous potassium and iodine molecules. As this process requires breaking ionic bonds, it is endothermic (螖H鈦 > 0). Moreover, the reaction results in an increase in the number of gas molecules, leading to an increase in entropy (螖S鈦 > 0).
04

(ii) Reaction 2: Predicting K > 1 and the Temperature Dependence of K

With both 螖H鈦 and 螖S鈦 being positive, the reaction is more spontaneous at higher temperatures, which means K > 1 at high temperatures. K increases with increasing temperature.
05

(iii) Reaction 3: Predicting 螖H鈦 and 螖S鈦

In this reaction, Na鈧(g) dissociates into 2Na(g). The reaction requires bond breaking, so it's endothermic (螖H鈦 > 0). Additionally, since the reaction increases the number of gas molecules, there is an increase in entropy (螖S鈦 > 0).
06

(iii) Reaction 3: Predicting K > 1 and the Temperature Dependence of K

Both 螖H鈦 and 螖S鈦 are positive for this reaction, which means the reaction is more spontaneous at higher temperatures, and K increases with an increase in temperature.
07

(iv) Reaction 4: Predicting 螖H鈦 and 螖S鈦

In this reaction, 2V鈧侽鈧(s) decomposes into 4V(s) and 5O鈧(g). The reaction requires bond breaking, so it's endothermic (螖H鈦 > 0). Furthermore, the reaction leads to an increase in the number of gas molecules, suggesting an increase in entropy (螖S鈦 > 0).
08

(iv) Reaction 4: Predicting K > 1 and the Temperature Dependence of K

As both 螖H鈦 and 螖S鈦 are positive for this reaction, the reaction is more spontaneous at higher temperatures (K > 1), and K increases with increasing temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change (螖H鈦)
Enthalpy change, denoted as 螖H鈦, represents the heat change in a reaction at constant pressure. It's a vital concept in determining whether a reaction is exothermic or endothermic.

For exothermic reactions, 螖H鈦 is negative, meaning heat is released to the surroundings. These reactions feel warm to touch and typically involve bond formation. Conversely, endothermic reactions have a positive 螖H鈦, which indicates heat absorption, often involving bond breaking.

Understanding the enthalpy change of a reaction can help predict its feasibility and energy efficiency. For instance, in reaction (i) with Mg and O鈧 forming MgO, 螖H鈦 is negative because forming ionic bonds releases energy. Remember, bond making releases energy, while bond breaking requires energy.
Entropy Change (螖S鈦)
Entropy change, symbolized as 螖S鈦, quantifies the disorder or randomness in a system during a chemical reaction. Nature tends to favor more disordered states, hence an increase in entropy usually leads to more spontaneous processes.

When a reaction results in more gas molecules or breakdown into smaller particles, 螖S鈦 is positive. This means the system's entropy increases. An example is reaction (ii), where solid KI decomposes into gaseous K and I鈧, increasing system randomness.

Conversely, if a reaction results in fewer gas molecules, 螖S鈦 is negative. This is seen in reaction (i) where MgO formation decreases the number of gas molecules, thus reducing entropy. Understanding 螖S鈦 helps determine reaction spontaneity.
Equilibrium Constant (K)
The equilibrium constant, denoted as K, provides insight into the position of equilibrium within a chemical reaction. It relates to the concentration of reactants and products at equilibrium.

For a given reaction at equilibrium, if K > 1, products are favored, meaning the reaction proceeds effectively in the forward direction. If K < 1, reactants are favored, indicating limited product formation.

In our example, reactions (ii), (iii), and (iv) have K > 1 at higher temperatures since they have both positive 螖H鈦 and 螖S鈦, making them more favorable thermodynamically at elevated temperatures. Understanding K and its dependence on temperature can predict how a reaction behaves under different conditions.
Temperature Dependence of Reactions
Temperature plays a crucial role in chemical reactions, influencing factors like reaction rate and equilibrium position. The interplay between enthalpy (螖H鈦) and entropy (螖S鈦) changes further dictates how temperature affects a reaction.

For reactions with both 螖H鈦 and 螖S鈦 positive, an increase in temperature leads to a larger equilibrium constant, making the reaction more spontaneous. This is why reactions such as (ii), (iii), and (iv) become more favorable at higher temperatures.

Alternatively, reactions with both 螖H鈦 and 螖S鈦 negative, like reaction (i), tend to have decreasing K with increasing temperature, reducing their spontaneity. Understanding these dynamics can guide predictions on how reactions shift with temperature changes.

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Most popular questions from this chapter

An ice cube with a mass of 20 \(\mathrm{g}\) at \(-20^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds 500 \(\mathrm{mL}\) of hot water, initially at \(83^{\circ} \mathrm{C} .\) What is the final temperature in the cup? The density of liquid water is 1.00 \(\mathrm{g} / \mathrm{mL}\) ; the specific heat capacity of ice is \(2.03 \mathrm{J} / \mathrm{g}-\mathrm{C}\) ; the specific heat capacity of liquid water is \(4.184 \mathrm{J} / \mathrm{g}-\mathrm{C} ;\) the enthalpy of fusion of water is 6.01 \(\mathrm{k} \mathrm{J} / \mathrm{mol} .\)

The element gallium (Ga) freezes at \(29.8^{\circ} \mathrm{C},\) and its molar enthalpy of fusion is \(\Delta H_{\text { fus }}=5.59 \mathrm{k} \mathrm{k} / \mathrm{mol}\) . (a) When molten gallium solidifies to Ga(s) at its normal melting point, is \(\Delta S\) positive or negative? (b) Calculate the value of \(\Delta S\) when 60.0 g of Ga(l) solidifies at \(29.8^{\circ} \mathrm{C}\) .

Consider the reaction $$ \mathrm{PbCO}_{3}(s) \rightleftharpoons \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) $$ Using data in Appendix C, calculate the equilibrium pres- sure of \(\mathrm{CO}_{2}\) in the system at (a) \(400^{\circ} \mathrm{C}\) and \((\mathbf{b}) 180^{\circ} \mathrm{C} .\)

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\mathrm{vap}}=29.6 \mathrm{kJ} / \mathrm{mol}\) (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when 1.00 mol of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8^{\circ} \mathrm{C}\) .

A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{kJ}\) and \(\Delta S^{\circ}=\) \(+52.4 \mathrm{J} / \mathrm{K}\) . (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system?(c) Calculate \(\Delta G^{\circ}\) for the reaction at 298 \(\mathrm{K}\) . (d) Is the reaction spontaneous at 298 \(\mathrm{K}\) under standard conditions?

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