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Chapter 9: Problem 81

Calculate the enthalpy change for the reaction \(\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\), from the following equations: (9.6) $$ \begin{array}{ll} \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g) & \Delta H=-92 \mathrm{~kJ} \\ \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NH}_{4} \mathrm{Cl}(s) & \Delta H=-631 \mathrm{~kJ} \\ \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) & \Delta H=-296 \mathrm{~kJ} \end{array} $$

Short Answer

Expert verified
The enthalpy change for the reaction is \(75.5 \mathrm{~kJ}\).

Step by step solution

01

Write the Target Equation

The target equation is given as \ \ \( \mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \). \ We need to find the enthalpy change for this reaction.
02

Identify the Given Equations and Their Enthalpy Changes

The provided equations are: \ \ (i) \( \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g) \) \ \( \Delta H=-92 \mathrm{~kJ} \) \ \ (ii) \( \mathrm{N}_{2}(g)+4 \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NH}_{4} \mathrm{Cl}(s) \) \ \( \Delta H=-631 \mathrm{~kJ} \) \ \ (iii) \( \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \) \ \( \Delta H=-296 \mathrm{~kJ} \)
03

Manipulate Equation (ii) to Isolate \( \mathrm{NH}_{4} \mathrm{Cl}(s) \)

We need 1 mole of \( \mathrm{NH}_{4} \mathrm{Cl}(s) \), so divide equation (ii) by 2: \ \ \( \frac{1}{2} \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s) \) \ \( \Delta H = - \frac{631}{2} \mathrm{~kJ} = -315.5 \mathrm{~kJ} \)
04

Reverse the Manipulated Equation (ii)

Reversing the equation to get \( \mathrm{NH}_{4} \mathrm{Cl}(s) \rightarrow \frac{1}{2} \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \) changes the enthalpy sign: \ \ \( \Delta H = +315.5 \mathrm{~kJ} \)
05

Multiply Equation (iii) to Match \( \mathrm{NH}_{3}(g) \)

We need 1 mole of \( \mathrm{NH}_{3}(g) \), so divide equation (iii) by 2: \ \ \( \frac{1}{2} \mathrm{N}_{2}(g) + \frac{3}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) \) \ \( \Delta H = - \frac{296}{2} \mathrm{~kJ} = -148 \mathrm{~kJ} \)
06

Write the Given Enthalpy for \( \mathrm{HCl}(g) \)

Use the given equation (i) directly: \ \ \( \frac{1}{2} \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g) \) \ \( \Delta H = -92 \mathrm{~kJ} \)
07

Combine All Manipulated Equations

Combine the updated enthalpy changes: \ \ \( \mathrm{NH}_{4} \mathrm{Cl}(s) \rightarrow \frac{1}{2} \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \) \ \( \Delta H = +315.5 \mathrm{~kJ} \) \ \ \( \frac{1}{2} \mathrm{N}_{2}(g) + \frac{3}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) \) \ \( \Delta H = -148 \mathrm{~kJ} \) \ \ \( \frac{1}{2} \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g) \) \ \( \Delta H = -92 \mathrm{~kJ} \)
08

Sum the Enthalpy Changes

Add the enthalpy changes from the manipulated equations: \ \ \( \Delta H = 315.5 + (-148) + (-92) = 315.5 - 240 = 75.5 \mathrm{~kJ} \)

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Key Concepts

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

enthalpy
Enthalpy is a measurement of the total energy of a thermodynamic system. It includes the internal energy of the system plus the product of its volume and pressure. In simpler terms, it is the heat content of a system at constant pressure. Enthalpy changes, denoted as \(\Delta H\), occur when a chemical reaction absorbs or releases heat.
Positive enthalpy change (\(\Delta H > 0\)) means heat is absorbed, making it an endothermic reaction. Negative enthalpy change (\(\Delta H < 0\)) means heat is released, which is an exothermic reaction.
chemical equations
A chemical equation represents a chemical reaction where the reactants are written on the left side and products on the right side, separated by an arrow. This arrow indicates the direction and type of the reaction.
For example, in the given problem: \( NH_{4}Cl(s) \rightarrow NH_{3}(g) + HCl(g) \)
Ammonium chloride (\

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Most popular questions from this chapter

When \(30.0 \mathrm{~g}\) of carbon is heated with silicon dioxide, \(28.2 \mathrm{~g}\) of carbon monoxide is produced. What is the percent yield of carbon monoxide for this reaction? $$ \mathrm{SiO}_{2}(s)+3 \mathrm{C}(s) \stackrel{\Delta}{\longrightarrow} \operatorname{SiC}(s)+2 \mathrm{CO}(g) $$

Carbon disulfide and carbon monoxide are produced when carbon is heated with sulfur dioxide. $$ 5 \mathrm{C}(s)+2 \mathrm{SO}_{2}(g) \stackrel{\Delta}{\longrightarrow} \mathrm{CS}_{2}(l)+4 \mathrm{CO}(g) $$ a. How many moles of \(\mathrm{C}\) are needed to react with \(0.500 \mathrm{~mol}\) of \(\mathrm{SO}_{2} ?\) b. How many moles of \(\mathrm{CO}\) are produced when \(1.2 \mathrm{~mol}\) of \(\mathrm{C}\) reacts? c. How many moles of \(\mathrm{SO}_{2}\) are required to produce \(0.50 \mathrm{~mol}\) of \(\mathrm{CS}_{2} ?\) d. How many moles of \(\mathrm{CS}_{2}\) are produced when \(2.5 \mathrm{~mol}\) of \(\mathrm{C}\) reacts?

Sulfur reacts with carbon to form carbon disulfide. \((9.2,9.6)\) $$ \mathrm{C}(s)+2 \mathrm{~S}(s) \longrightarrow \mathrm{CS}_{2}(g) \quad \Delta H=+92 \mathrm{~kJ} $$ a. Is the reaction endothermic or exothermic? b. How many kilojoules are required when \(1.5 \mathrm{~mol}\) of S reacts? c. How many kilojoules are required when \(100 \mathrm{~g}\) of \(\mathrm{CS}_{2}\) is formed?

The equation for the reaction of iron and oxygen gas to form rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) is written as \((9.2,9.6)\) $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1.7 \times 10^{3} \mathrm{~kJ} $$ a. How many kilojoules are released when \(2.00 \mathrm{~g}\) of Fe reacts? b. How many grams of rust form when \(150 \mathrm{~kJ}\) are released?

Aluminum and chlorine combine to form aluminum chloride. $$ \begin{aligned} &(9.2,9.3,9.4,9.5) \\ &2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s) \end{aligned} $$ a. How many moles of \(\mathrm{Cl}_{2}\) are needed to react with \(4.50 \mathrm{~mol}\) of Al? b. How many grams of \(\mathrm{AlCl}_{3}\) are produced when \(50.2 \mathrm{~g}\) of Al reacts? c. When \(13.5 \mathrm{~g}\) of \(\mathrm{Al}\) reacts with \(8.00 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), how many grams of \(\mathrm{AlCl}_{3}\) can form? d. If \(45.0 \mathrm{~g}\) of \(\mathrm{Al}\) and \(62.0 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\) are mixed, and \(66.5 \mathrm{~g}\) of \(\mathrm{AlCl}_{3}\) is actually obtained, what is the percent yield of \(\mathrm{AlCl}_{3}\) for the reaction?
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