Derive the relationship between ∆ Ur and∆ Hr for reactions involving gaseousreactants and/or products
Question
Derive the relationship between and for reactions involving gaseous reactants and/or products
Solution
To derive the relationship between ∆Ur and ∆Hr for reactions involving gaseous reactants and/or products, we can use the equation:
∆Hr = ∆Ur + ∆(nRT)
where ∆Hr is the change in enthalpy of the reaction, ∆Ur is the change in internal energy of the reaction, ∆(nRT) is the change in the number of moles of gas multiplied by the ideal gas constant (R) and the temperature (T).
Here are the steps to derive this relationship:
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Start with the definition of enthalpy change (∆Hr) for a reaction: ∆Hr = H(products) - H(reactants) where H is the enthalpy of the system.
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Use the first law of thermodynamics to express the change in internal energy (∆Ur) in terms of the heat (q) and work (w) done on or by the system: ∆Ur = q - w
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For a reaction at constant pressure, the heat (q) is equal to the enthalpy change (∆Hr): q = ∆Hr
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The work done (w) can be expressed as the product of the change in volume (∆V) and the pressure (P): w = -P∆V
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For gaseous reactions, the change in volume (∆V) can be related to the change in the number of moles of gas (∆n) using the ideal gas law: ∆V = ∆nRT/P where ∆n is the difference in the number of moles of gas between products and reactants, R is the ideal gas constant, and T is the temperature.
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Substitute the expressions for q and w into the equation for ∆Ur: ∆Ur = ∆Hr - (-P∆V) ∆Ur = ∆Hr + P∆V
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Substitute the expression for ∆V from step 5 into the equation for ∆Ur: ∆Ur = ∆Hr + P(∆nRT/P) ∆Ur = ∆Hr + ∆nRT
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Simplify the equation by canceling out the pressure terms: ∆Ur = ∆Hr + ∆nRT
Therefore, the relationship between ∆Ur and ∆Hr for reactions involving gaseous reactants and/or products is given by the equation: ∆Hr = ∆Ur + ∆(nRT)
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