(a) For each of the following reactions, predict the sign of \(\Delta H^{*}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\) at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{FeO}(s)\) (ii) \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (iii) $\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$ (iv) $\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CaO}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)$

Step by step solution

01

Since this is a formation reaction where we are producing a compound from its elements, the enthalpy change is likely to be negative, meaning the reaction is exothermic. ##Step 2: Determine sign of ΔS°##

In this reaction, we have one mole of a gaseous reactant forming two moles of solid product. This means a decrease in entropy as gas has more entropy than solids. So, ΔS° would be negative. ##Step 3: Predict K > 1 at 25°C##

02

Given that ΔH* is negative and ΔS° is negative, the behavior of the reaction depends on the relative magnitudes of the two values. However, typically, formation reactions tend to proceed spontaneously, so we can say that K > 1 at 25°C. ##Step 4: Dependence of K on temperature##

According to Van't Hoff's equation(\( \frac{d \ln K }{ dT } = \frac{ \Delta H }{ RT^2 } \)), a reaction with negative ΔH* and negative ΔS° will have K decrease with increasing temperature. #Reaction 2: Cl₂(g) ⇌ 2 Cl(g)# ##Step 1: Determine sign of ΔH*##

03

This reaction involves breaking the Cl-Cl bond in Cl₂, which requires energy input. Therefore, the enthalpy change will be positive, meaning the reaction is endothermic. ##Step 2: Determine sign of ΔS°##

The reaction involves one mole of reactant gas forming two moles of product gas. As a result, there is an increase in entropy in the system. Therefore, ΔS° is positive. ##Step 3: Predict K > 1 at 25°C##

04

As both ΔH* and ΔS° are positive, the reaction will tend to be more spontaneous at higher temperatures. At 25°C, we cannot predict with certainty whether or not K > 1 without more information. ##Step 4: Dependence of K on temperature##

A reaction with positive ΔH* and positive ΔS° will have K increase with increasing temperature according to Van't Hoff's equation. #Reaction 3: NH₄Cl(s) ⇌ NH₃(g) + HCl(g)## ##Step 1: Determine sign of ΔH*##

05

This reaction involves breaking the ionic bond in NH₄Cl and forming two moles of gas, so energy input is required. The enthalpy change is positive, meaning the reaction is endothermic. ##Step 2: Determine sign of ΔS°##

Going from one mole of a solid substance to two moles of gases results in an increase in entropy. Therefore, ΔS° is positive. ##Step 3: Predict K > 1 at 25°C##

06

With positive ΔH* and ΔS°, this reaction will be favored at higher temperatures, making it difficult to predict whether K > 1 at 25°C. ##Step 4: Dependence of K on temperature##

A reaction with positive ΔH* and positive ΔS° will have K increase with increasing temperature according to Van't Hoff's equation. #Reaction 4: CO₂(g) + CaO(s) ⇌ CaCO₃(s)## ##Step 1: Determine sign of ΔH*##

07

This is a formation reaction, so the enthalpy change is likely to be negative, meaning the reaction is exothermic. ##Step 2: Determine sign of ΔS°##

One mole of gas and one mole of solid are converted into one mole of solid product, leading to a decrease in entropy as a gas has more entropy than solids. Therefore, ΔS° is negative. ##Step 3: Predict K > 1 at 25°C##

08

Generally, in formation reactions, K tends to be greater than 1. However, we cannot predict with certainty without more information on the relative magnitudes of ΔH* and ΔS°. ##Step 4: Dependence of K on temperature##

A reaction with negative ΔH* and negative ΔS° will have K decrease with increasing temperature according to Van't Hoff's equation.

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