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Chapter 12: Problem 21

Calculate the standard free energy change for the reaction, $$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \text { at } 298 \mathrm{~K} $$ Given \(\Delta H^{\circ}=-92.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-198.3 \mathrm{JK}^{-1} .\) Also, comment on the result.

Short Answer

Expert verified
\(\Delta G^\circ = -33.3 \mathrm{~kJ}\). The reaction is spontaneous at 298 K.

Step by step solution

01

Understanding the Gibbs Free Energy equation

The standard free energy change (\(\Delta G^\circ\)) for a chemical reaction at a constant temperature can be calculated using the Gibbs Free Energy equation: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\), where \(\Delta H^\circ\) is the standard change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S^\circ\) is the standard change in entropy.
02

Converting given values to consistent units

Ensure that all values are in appropriate units for the Gibbs Free Energy equation. Here, \(\Delta H^\circ\) needs to be in joules so we convert it: \(\Delta H^\circ = -92.4 \mathrm{~kJ} = -92,400 \mathrm{~J}\). The temperature is already given in Kelvin and \(\Delta S^\circ\) is already in joules per kelvin, so no further conversion is needed.
03

Plugging in values to the Gibbs Free Energy equation

Insert the known values into the Gibbs Free Energy equation: \(\Delta G^\circ = (-92,400 \mathrm{~J}) - (298 \mathrm{~K})(-198.3 \mathrm{~J K^{-1}})\).
04

Calculating \(\Delta G^\circ\)

Perform the calculations to find the standard free energy change: \(\Delta G^\circ = -92,400 \mathrm{~J} + 59,092.4 \mathrm{~J} = -33,307.6 \mathrm{~J} = -33.3 \mathrm{~kJ}\) (since \(1 \mathrm{~J} = 0.001 \mathrm{~kJ}\)).
05

Commenting on the result

The negative value of \(\Delta G^\circ\) indicates that the reaction is spontaneous under standard conditions at 298 K.

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as ewlineewlineewlineewline The temperature during the reaction must be specified, most commonly represented in Kelvin (K). In the given exercise, the enthalpy change for the formation of ammonia from nitrogen and hydrogen gases (ewlineewlineewlineewline To fully understand how enthalpy changes impact reactions, it's crucial to consider not only the absolute value but also the sign: a negative ewlineewlineewlineewline
Entropy Change
Entropy refers to the degree of randomness or disorder in a system, and changes in entropy (ewlineewlineewlineewline In a chemical reaction, a positive ewlineewlineewlineewline In the provided exercise, the entropy change for the reaction is given as negative, ewlineewlineewlineewline
Spontaneous Reaction
A spontaneous reaction is one that occurs naturally without the need for external energy once it has been initiated. Determining if a reaction is spontaneous involves understanding the combined effects of enthalpy and entropy changes on the Gibbs Free Energy. For a reaction to be spontaneous at a constant temperature and pressure, the Gibbs Free Energy change (ewlineewlineewlineewline In the exercise, we calculated a negative ewlineewlineewlineewline
Chemical Thermodynamics
Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions within the boundaries of thermodynamic systems. It's governed by several laws and principles, such as the first and second laws of thermodynamics. Key concepts within this field include enthalpy, entropy, free energy, and the equilibrium constant, which provide insights into reaction spontaneity and the energy changes during a reaction.ewlineewlineewlineewline When we combine the enthalpy, entropy, and Gibbs Free Energy for reactions, as in our exercise, we're utilizing core principles of chemical thermodynamics to describe and predict the behavior of chemical systems.
Standard Free Energy Change
The standard free energy change (ewlineewlineewlineewline It serves as a predictive tool: negative values signal spontaneous reactions under standard conditions, zero indicates equilibrium, and positive values designate non-spontaneous reactions unless influenced by external energy. The exercise provided a context to this concept, wherein the calculated negative value of ewlineewlineewlineewline

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Most popular questions from this chapter

Calculate the entropy change for the conversion of following: (a) \(1 \mathrm{~g}\) ice to water at \(273 \mathrm{~K}, \Delta H_{\mathrm{f}}\) for ice \(=6.025 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (b) \(36 \mathrm{~g}\) water to vapour at \(373 \mathrm{~K} ; \Delta H_{\mathrm{v}}\) for \(\mathrm{H}_{2} \mathrm{O}=40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

For the water gas reaction : $$ \mathrm{C}_{(\mathrm{s})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2(\mathrm{~g})} $$ the standard Gibbs energy of reaction (at \(1000 \mathrm{~K}\) ) is \(-8.1 \mathrm{~kJ} \mathrm{~mol}^{-\mathrm{i}}\). Calculate its equilibrium constant.

During \(200 \mathrm{~J}\) work done on the system, \(140 \mathrm{~J}\) of heat is given out. Calculate the change in internal energy.

Two litre of \(\mathrm{N}_{2}\) at \(0^{\circ} \mathrm{C}\) and 5 atm pressure are expanded isothermally against a constant external pressure of 1 atm until the pressure of gas reaches 1 atm. Assuming gas to be ideal, calculate work of expansion.

Calculate the equilibrium constant \(K_{\mathrm{p}}\) for the reaction given below if \(\Delta G^{\circ}=-10.632 \mathrm{~kJ}\) at \(300 \mathrm{~K}\) $$ \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} $$
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