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The amount of heat evolved when \(0.50 \mathrm{~m}_{0} \mathrm{e}\) of \(\mathrm{HCl}\) is mixed with \(0.30\) mole of \(\mathrm{NaOH}\) solution is (a) \(57.1 \mathrm{~kJ}\) (b) \(28.55 \mathrm{~kJ}\) (c) \(11.42 \mathrm{~kJ}\) (d) \(17.13 \mathrm{~kJ}\)

Short Answer

Expert verified
17.13 kJ

Step by step solution

01

- Identify the Reaction

Recognize that when HCl reacts with NaOH, it forms a neutralization reaction which results in the formation of water and sodium chloride. The chemical equation for this reaction is: \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \).
02

- Reaction Stoichiometry

From the balanced chemical equation, we can see that the reaction consumes equal moles of HCl and NaOH to produce water and sodium chloride. Since we have excess HCl (0.50 mole) compared to NaOH (0.30 mole), NaOH is the limiting reagent.
03

- Calculate Heat Evolved

The heat evolved during the reaction is proportional to the amount of the limiting reagent. The reaction of 1 mole of NaOH with HCl releases 57.1 kJ of heat. Therefore, the heat evolved from 0.30 mole of NaOH is: \( 0.30 \text{ mol} \times 57.1 \frac{\text{kJ}}{\text{mol}} = 17.13 \text{kJ} \).

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Key Concepts

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

Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt, usually accompanied by the evolution of heat. Acids and bases, according to the Brønsted-Lowry definition, are substances that can give up and accept protons (H+ ions), respectively. In our exercise instance, hydrochloric acid (HCl) donates a proton to sodium hydroxide (NaOH), which accepts it, thereby 'neutralizing' each other's properties.

The resultant chemical equation for the reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \].

This type of reaction is exothermic, releasing energy in the form of heat, which is a significant concept when understanding heat evolved in chemical reactions. The actual amount of heat released is influenced by the specific substances involved and the quantities that react. From practical experience, such reactions are common in laboratory settings for determining the concentrations of acid or base solutions, highlighting their real-world relevance.
Reaction Stoichiometry
Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculations based upon the balanced chemical equation, allowing one to predict the amounts of reactants needed or products formed.

In our particular problem, the balanced equation shows a one-to-one molar ratio between HCl and NaOH:
\[ \text{1 mol HCl} + \text{1 mol NaOH} \rightarrow \text{1 mol NaCl} + \text{1 mol H}_2\text{O} \].

This ratio is crucial for predicting how much product will form. It also helps us understand one of the pivotal concepts of stoichiometry: the limiting reagent. This is especially important when the reactants are provided in non-stoichiometric amounts, as is the case in the exercise. Correct stoichiometric calculations are foundational not only in academic settings but are also essential in industrial processes where efficiency and cost-effectiveness are paramount.
Limiting Reagent
The limiting reagent in a chemical reaction is the substance that is entirely consumed first, limiting the extent of the reaction and determining the amount of product formed. Once the limiting reagent is used up, the reaction cannot proceed further, even if other reactants are still present in excess.

In the exercise, we have 0.50 moles of HCl and 0.30 moles of NaOH. Since the balanced chemical equation shows that HCl and NaOH react in a 1:1 molar ratio, the smaller amount of NaOH will be the limiting reagent, as it will run out before HCl. Understanding which reagent is limiting is a critical skill in chemistry, as it lets us accurately calculate theoretical yields and can help manage resources more effectively in commercial chemical production. For our exercise, knowing that NaOH is the limiting reagent allows us to focus only on its quantity when calculating the total heat evolved during the neutralization.

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Most popular questions from this chapter

A system absorbs \(50 \mathrm{~kJ}\) heat and does \(20 \mathrm{~kJ}\) of work. What is the net change in the internal energy of the system? (a) Increase by \(30 \mathrm{~kJ}\) (b) Decrease by \(30 \mathrm{~kJ}\) (c) Increase by \(70 \mathrm{~kJ}\) (d) Decrease by \(70 \mathrm{~kJ}\)

The enthalpy of solution of sodium chloride is \(4 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and its enthalpy of bydration of ions is \(-784\) kJ mol \(^{-1}\). What will be the lattice enthalpy of sodium chloride? (a) \(+780 \mathrm{k} J \mathrm{~mol}^{-1}\) (b) \(+394 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+788 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+398 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

What will be the work done when one mole of a gas expands isothermally from \(15 \mathrm{~L}\) to \(50 \mathrm{~L}\) against a constant pressure of 1 atm at \(25^{\circ} \mathrm{C}\) ? (a) \(-3542 \mathrm{cal}\) (b) \(-843.3 \mathrm{cal}\) (c) \(-718 \mathrm{cl}\) (d) \(-60.23 \mathrm{cal}\)

Match the column I with column II and mark the appropriate choice. $$\begin{array}{|l|l|l|l|} \hline {\text { Column I }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { State function } & \text { (i) } & \text { At constant pressure } \\ \hline \text { (B) } & \Delta H=q & \text { (ii) } & \text { Specific heat } \\\ \hline \text { (C) } & \Delta U=q & \text { (iii) } & \text { Entropy } \\ \hline \text { (D) } & \text { Intensive property } & \text { (iv) } & \text { At constant volume } \\ \hline \end{array}$$

What will be the change in internal energy when 12 kJ of work is done on the system and 2 kJ of heat is given by the system? (a) \(+10 \mathrm{~kJ}\) (b) \(-10 \mathrm{~kJ}\) (c) +5kJ (d) \(-5 \mathrm{~kJ}\)

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