Based on the equation and the information in the table, what is the enthalpy of the reaction?
[tex]\( \Delta H_{xn}=4619 kJ\)[/tex]
[tex]\( \Delta H_{xn}=-46.19 kJ\)[/tex]
[tex]\( \Delta H_{rxn}=92.38 kJ\)[/tex]
[tex]\( \Delta H_{rxn}=-92.38 kJ\)[/tex]
Answer (2)
The problem requires determining the enthalpy of a reaction without knowing the reaction itself.
Analyze the given options and relate them to the provided standard enthalpies of formation.
Recognize that 92.38 kJ is twice the absolute value of the enthalpy of formation of N H 3 ( g ) .
Conclude that 92.38 k J is the most plausible answer based on the given data.
Explanation
Problem Analysis We are given a table of standard enthalpies of formation for various substances and asked to determine the enthalpy of a reaction based on the provided options. However, the reaction itself is not provided. We must choose the most plausible option from the given values.
Strategy Since we don't have the reaction equation, we cannot calculate the enthalpy of the reaction directly using the standard enthalpies of formation. Instead, we look for a value among the options that seems related to the values in the table. The options are 4619 kJ, -46.19 kJ, 92.38 kJ, and -92.38 kJ.
Calculations and Reasoning We notice that -46.19 kJ is the standard enthalpy of formation for N H 3 ( g ) . Also, 92.38 kJ is exactly twice the absolute value of -46.19 kJ, which is the enthalpy of formation of N H 3 ( g ) . It is also close to the absolute value of the enthalpy of formation of H Cl ( g ) , which is -92.30 kJ. The value -92.38 kJ is the negative of 92.38 kJ.
Selecting the Answer Since 92.38 kJ is twice the magnitude of the enthalpy of formation of N H 3 ( g ) , it is a likely candidate if the reaction involves 2 moles of N H 3 ( g ) as either a reactant or a product. Also, -92.38 kJ is twice the enthalpy of formation of N H 3 ( g ) .
Final Answer Without the actual reaction, we can only make an educated guess. Given that -46.19 kJ is directly from the table and 92.38 kJ is a simple multiple (2 times) of a value in the table, we choose the option that is a multiple of a value in the table.
Examples
Enthalpy changes are crucial in designing chemical processes. For example, in designing an industrial ammonia synthesis plant, understanding the enthalpy of formation of ammonia ( N H 3 ) is essential to optimize reaction conditions and energy efficiency. If the reaction involves multiple moles of ammonia, the overall enthalpy change will be a multiple of the standard enthalpy of formation, directly impacting the energy requirements and cost of production. This knowledge helps engineers to create sustainable and economically viable processes.
The enthalpy of the reaction is most plausibly 92.38 kJ , as it is twice the absolute value of the standard enthalpy of formation of a commonly known compound, indicating a relationship in the reaction. This choice suggests a reaction involving two moles of that substance. Therefore, 92.38 kJ is the correct answer.
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