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Chapter 6: Problem 49

How much heat is evolved if \(3.2 \mathrm{~g}\) of methane is burnt and if the heat of combustion of methane is \(-880 \mathrm{~kJ} \mathrm{~mol}^{-1} ?\) (a) \(88 \mathrm{~kJ}\) (b) \(264 \mathrm{~kJ}\) (c) \(176 \mathrm{~kJ}\) (d) \(440 \mathrm{~kJ}\)

Short Answer

Expert verified
The heat evolved is 264 kJ.

Step by step solution

01

Calculate the number of moles of methane

Convert the mass of methane to moles using the molar mass of methane (CH4), which is approximately 16 g/mol. Number of moles = mass (g) / molar mass (g/mol).
02

Calculate the heat evolved

Multiply the number of moles of methane by the heat of combustion per mole. The heat evolved will be in kJ if the heat of combustion is given in kJ/mol.
03

Confirm the answer

Ensure the answer makes sense physically and corresponds to one of the given options.

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Key Concepts

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

Enthalpy Change
Enthalpy change, represented by ΔH, is a measure of the total heat content change in a chemical reaction at constant pressure.
When a substance combusts, the heat released during the reaction is known as the heat of combustion, which is a specific type of enthalpy change.
This value is often expressed in kilojoules per mole (kJ/mol), indicating the energy change associated with the combustion of one mole of a substance.
In an exothermic reaction like combustion, the enthalpy change is negative, signifying that energy is released to the surroundings.
Understanding the concept of enthalpy change is essential for grasping the amount of energy involved in chemical reactions.
Stoichiometry
Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction.
By using a balanced chemical equation, one can determine how much product can be formed from given masses of reactants or vice versa.
For combustion reactions, the stoichiometry allows us to calculate the amount of heat evolved based on the mole ratio between the fuel and oxygen and between the fuel and the energy produced.
It is imperative in predicting the outcomes of reactions and is a foundational tool when dealing with exercises involving chemical reactions and energy changes.
Molar Mass
Molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance.
The molar mass is typically expressed in units of grams per mole (g/mol), and it corresponds to the mass of one mole of a substance.
Knowing the molar mass allows us to convert between the mass of a substance and the number of moles, a key step in stoichiometric calculations.
Molar mass is particularly important when quantifying the reactants and products in chemical reactions, such as determining how much heat is produced during the combustion of a specific mass of fuel.
Chemical Thermodynamics
Chemical thermodynamics involves the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics.
It provides a thorough framework that explains how energy is transformed and conserved during chemical processes.
In the context of combustion, chemical thermodynamics deals with how the energy stored in chemical bonds is released as heat.
It also addresses how enthalpy change, a fundamental concept of thermodynamics, is applied to calculate the heat evolved in chemical reactions.
Understanding chemical thermodynamics is crucial for predicting the feasibility and spontaneity of reactions, as well as for designing processes that efficiently convert chemical energy into useful work.

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Most popular questions from this chapter

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) \(-5 \mathrm{~kJ}\)

Match the following columns and mark the appropriate choice. \(\begin{array}{|l|l|l|l|} \hline {\text { Column I }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { Exothermic } & \text { (i) } & \Delta H=0, \Delta E=0 \\ \hline \text { (B) } & \text { Spontaneous } & \text { (ii) } & \Delta G=0 \\ \hline \text { (C) } & \text { Cyclic process } & \text { (iii) } & \Delta H \text { is negative. } \\ \hline \text { (D) } & \text { Equilibrium } & \text { (iv) } & \Delta G \text { is negative. } \\ \hline \end{array}\) (a) \((\mathrm{A}) \rightarrow(\mathrm{ii}),(\mathrm{B}) \rightarrow(\mathrm{iii}),(\mathrm{C}) \rightarrow(\mathrm{i}),(\mathrm{D}) \rightarrow(\mathrm{iv})\) (b) \((\mathrm{A}) \rightarrow(\mathrm{iv}),(\mathrm{B}) \rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow(\mathrm{iii}),(\mathrm{D}) \rightarrow\) (ii) (c) (A) \(\rightarrow\) (i), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) \((A) \rightarrow(i i i),(B) \rightarrow(i v),(C) \rightarrow(i),(D) \rightarrow(i i)\)

What will be the signs of \(\Delta H\) and \(\Delta S\) when \(\mathrm{NaOH}\) is dissolved in water? \(\begin{array}{ll}\Delta H & \Delta S \\ \text(a) - & \- \\ \text(c) - & \+ & \end{array}\) \(\begin{array}{ll}\Delta H & \Delta S \\ \text(b) + & \- \\ \text(d) + & +\end{array}\)

In a reaction \(P+Q \rightarrow R+S\), there is no change in entropy. Enthalpy change for the reaction \((\Delta H)\) is \(12 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Under what conditions, reaction will have negative value of free energy change? (a) If \(\Delta H\) is positive. (b) If \(\Delta H\) is negative. (c) If \(\Delta H\) is \(24 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (d) If temperature of reaction is high.

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