The integral enthalpy of solution in \(\mathrm{kJ}\) of one mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(n\) mole of water is given by: $$ \Delta H_{s}=\frac{75.6 \times n}{n+1.8} $$ Calculate \(\Delta H\) for the following process: (a) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(2 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\). (b) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in 7 mole of \(\mathrm{H}_{2} \mathrm{O}\). (c) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in 5 mole of \(\mathrm{H}_{2} \mathrm{O}\). (d) solution (a) dissolved in 5 mole of \(\mathrm{H}_{2} \mathrm{O}\). (e) 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in excess of \(\mathrm{H}_{2} \mathrm{O}\).

Short Answer

Expert verified
\(\Delta H_s\) values: (a) 50.4 kJ, (b) 72.6 kJ, (c) 67.7 kJ, (d) 72.6 kJ, (e) 75.6 kJ

Step by step solution

01

Calculate \(\Delta H_s\) for solution (a)

Use the given formula substituting \(n = 2\text{ moles}\) of \(\text{H}_2\text{O}\) to find \(\Delta H_s\) for 1 mole of \(\text{H}_2\text{SO}_4\) dissolved in 2 moles of water.
02

Calculate \(\Delta H_s\) for solution (b)

Use the given formula substituting \(n = 7\text{ moles}\) of \(\text{H}_2\text{O}\) to find \(\Delta H_s\) for 1 mole of \(\text{H}_2\text{SO}_4\) dissolved in 7 moles of water.
03

Calculate \(\Delta H_s\) for solution (c)

Use the given formula substituting \(n = 5\text{ moles}\) of \(\text{H}_2\text{O}\) to find \(\Delta H_s\) for 1 mole of \(\text{H}_2\text{SO}_4\) dissolved in 5 moles of water.
04

Calculate \(\Delta H_s\) for solution (d)

Note that solution (a) describes the dissolution in 2 moles of \(\text{H}_2\text{O}\), but then it's further dissolved in 5 moles of water, making the total \(n = 2 + 5 = 7\) moles. Use the formula with \(n = 7\) moles.
05

Calculate \(\Delta H_s\) for solution (e) when \(\text{H}_2\text{O}\) is in excess

As \(n\) approaches infinity (excess of water), the fraction \(\frac{n}{n+1.8}\) approaches 1. So, \(\Delta H_s\) approaches the constant factor 75.6 \(\text{kJ/mole}\).

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemistry, it often involves the study of energy changes and transfers that occur during chemical reactions and changes of state.

Enthalpy, denoted by the symbol \(H\), is one of the primary thermodynamic quantities used to describe the heat energy involved in chemical processes. When a solute dissolves in a solvent, we encounter the concept of enthalpy of solution, \(\Delta H_s\), which represents the heat change when one mole of solute dissolves completely in a solvent. According to the First Law of Thermodynamics, energy cannot be created or destroyed in an isolated system, hence the energy released or absorbed in solution formation must equal the energy change of the system.

For students mastering IIT-JEE chemistry problems, understanding the calculations of \(\Delta H_s\) is critical, as it involves applying the thermodynamic principle of conservation of energy to practical, real-world chemical scenarios.
Physical Chemistry
Physical chemistry bridges the gap between the theories of physics and the reactions of chemistry, and it deals with the physical structure of chemical compounds and the way they react. In solution chemistry, it specifically looks at the interactions and energy changes that occur when substances dissolve.

The dissolution process can be highly endothermic or exothermic, depending on the nature of the solute and solvent, as well as their interactions. The integral enthalpy of solution, the exercise's focal point, is a concept from physical chemistry that measures the heat change under constant pressure when a substance dissolves in a solvent.

Understanding the maths and physics behind enthalpy changes in solutions not only prepares students for exams like the IIT-JEE but also for practical laboratory work where such concepts are frequently applied.
IIT-JEE Chemistry Problems
The IIT-JEE (Indian Institutes of Technology Joint Entrance Examination) is a highly competitive exam that tests a student's understanding of core science concepts, including chemistry. Problems related to thermodynamics, including the calculation of enthalpy of solution, often appear in the chemistry section of this exam.

To excel in these areas, students must have a solid understanding of solution chemistry and be proficient in applying the formulas to calculate various thermodynamic quantities. The step by step solution provided in the textbook aids in breaking down complex processes into manageable calculations - a skill that is vital for tackling IIT-JEE chemistry problems efficiently.

Moreover, practicing such problems helps in developing a methodical approach to problem-solving, which can greatly improve performance in the actual exam, where time management is as crucial as the understanding of concepts.
Solution Chemistry
Solution chemistry is a sub-branch of physical chemistry concerned with the study of solutions, which are homogeneous mixtures composed of a solute dissolved in a solvent. The enthalpy of solution is a key concept in this field, as it reflects the heat absorbed or released during the dissolution process.

Each substance has characteristic interactions with water, also known as the 'solvent of life', due to its incredible ability to dissolve so many substances. These interactions can be understood through the enthalpy change (\(\Delta H_s\)) when a substance, such as \(\text{H}_2\text{SO}_4\), dissolves in water. The study of these interactions is essential for students not only for academic exams like IIT-JEE but also for real-world applications such as drug formulation in pharmaceuticals or the analysis of pollutants in environmental chemistry.

When teaching the enthalpy of solution, it's important to emphasize that this thermodynamic property can predict the favorability of a given dissolution process and provide insight into the molecular interactions at play within the solution.

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Most popular questions from this chapter

The commercial production of water gas utilizes the reaction under standard conditions: \(\mathrm{C}+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}+\mathrm{CO}\). The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to \(\mathrm{CO}_{2}\). How many \(\mathrm{g}\) of carbon must be burnt to \(\mathrm{CO}_{2}\) to provide enough heat for the water gas conversion of \(100 \mathrm{~g}\) carbon? Neglect all heat losses to the environment. Also \(\Delta H_{\mathrm{f}}^{2}\) of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) and \(\mathrm{CO}_{2}\) are \(-110.53\), \(-241.81\) and \(-393.51 \mathrm{~kJ} / \mathrm{mol}\) respectively.

The heat of solution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in water was determined by measuring the amount of electrical work needed to compensate for the cooling which would otherwise óccur when the salt dissolves. After the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) was added to the water, electrical energy was provided by passage of current through a resistance coil until the temperature of the solution reached the value it had prior to the addition of salt. In a typical experiment, \(4.4 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) was added to \(200 \mathrm{~g}\) water. A current of \(0.75\) ampere was provided through the heater coil, and the voltage across the terminals was \(6.0 \mathrm{~V}\). The current was applied for \(5.2\) minute. Calculate \(\Delta H\) for the solution of \(1.0\) mole \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in enough water to give same concentration as was attained in the above experiment.

The standard enthalpy of formation of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is \(-65 \mathrm{kcal}\) \(\mathrm{mol}^{-1}\) and \(-197\) kcal mol \(^{-1}\) respectively. A mixture of two oxides contains \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the mole ratio \(2: 1 .\) If by oxidation, it is changed into a \(1: 2\) mole ratio mixture, how much of thermal energy will be released per mol of initial mixture.

\begin{aligned} &\text { The heat of dissociation of } \mathrm{H}_{2} \text { is } 435 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { . If } \mathrm{C}_{(\mathrm{s})} \longrightarrow \mathrm{C}_{(\mathrm{g})} \text { ; }\\\ &\Delta H=720 \mathrm{~kJ} \mathrm{~mol}^{-1}, \text { calculate the bond energy per mol of the } \mathrm{C}-\mathrm{H} \text { bond }\\\ &\text { in } \mathrm{CH}_{4} \text { molecule. } \Delta H_{\mathrm{f}} \text { for } \mathrm{CH}_{4}=-75 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { . } \end{aligned}

\(1.00\) litre sample of a mixture of \(\mathrm{CH}_{4(\mathrm{~g})}\) and \(\mathrm{O}_{2(\mathrm{~g})}\) measured at \(25^{\circ} \mathrm{C}\) and 740 torr was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of \(1260 \mathrm{cal} / \mathrm{K}\). The complete combustion of the methane to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) caused a temperature rise in the calorimeter of \(0.667 \mathrm{~K}\). What was the mole per cent of \(\mathrm{CH}_{4}\) in the original mixture? \(\Delta H_{\text {comb }}^{\circ}\left(\mathrm{CH}_{4}\right)=-215 \mathrm{k}\) cal \(\mathrm{mol}^{-1}\).

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