enthalpy H i the sum of the internal energy U and the 'pV-energy'= p*V.
The pV term is also called flow-energy, pressure-energy or energy of displacement. The p is the external pressure, but for systems at constant pressure it is also equal to the internal pressure. Hence: H=U+pV.
It follows that H is a composite form of energy, a statement not commonly found in the literature.
>> 'Energy-balance of the Joule-Thomson experiment: Enthalpy change at decompression.' (NPT-procestechnologie, 2010, 17(4)18-22)
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A negative enthalpy of formation indicates that energy is evolved.
I think you're going to have to give us the equation first, dear
The 1st step would be to give an example of the equation to be solved.
If you have two equations give AND one parametric equation why do you need to find yet another equation?
To solve Hess's law problems, first write out the chemical equations for all reactions involved. Then calculate the enthalpy change for each reaction. Finally, add or subtract the enthalpy changes to obtain the overall enthalpy change for the desired reaction.
The equation for ∆G is ∆G = ∆H - T∆S H is enthalpy and S is entropySo, ∆G is negative if T∆S is greater than ∆H
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"Try as he might, he could not solve the equation."
Delta H represents the change in enthalpy of a system. In the equation ΔG = ΔH - TΔS, it is the enthalpy change of the system. It indicates the heat absorbed or released during a reaction at constant pressure.
The enthalpy change to burn 37.5 g of ammonia (NH3) can be calculated using the standard enthalpy of formation of ammonia and the balanced chemical equation for its combustion. The enthalpy change will depend on the specific conditions of the reaction, such as temperature and pressure.
The enthalpy of formation equation for Ethanol (CH3CH2OH) can be written as: CH3CH2OH (l) -> C2H5OH (l) + 3/2 O2 (g) This equation represents the formation of 1 mol of Ethanol from its elements in their standard states at 25°C and 1 atm pressure.
One example of a chemical equation that includes the enthalpy change is the combustion of methane: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -802.3 kJ. In this reaction, 802.3 kJ of heat is released as the products are formed.
Heat appears in the equation as either a reactant (if heat is added to the reaction) or as a product (if heat is released by the reaction). It is typically denoted by the symbol "ΔH" for the change in enthalpy.
The difference between the enthalpy of formation of the products minus the enthalpy of formation of the reactants is the enthalpy of the reaction
For delta G to become negative at a given enthalpy and entropy, the process must be spontaneous. This can happen when the increase in entropy is large enough to overcome the positive enthalpy, leading to a negative overall Gibbs free energy. This typically occurs at higher temperatures where entropy effects dominate.
Specific Heat is:-the amount of energy required to raise the temperature of 1g of a substance 1°C-specific heat is symbolised as Cp or C-has units of J/g °C-J stands for Joules, which is a unit of energyMost often used in equation: q = mΔTCpq = energy in Jm = mass in gΔT = change in temperatureCp = specific heatExampleHow much energy is required to raise the temperature of 345.34g of Aluminium from 35.0°C to 250.00°C ? The specific heat of Aluminium is .90 J/Cp q = mΔTCpq = 345.34g(250.00°C - 35.0°C)0.09J/g°Cfrom there its just simple Algebra.q = 66,823.29JHowever, not all problems will be as simple as the one above. Some may require up to 3+ equations that can include Enthalpy or ΔH which leads to using a Change of State Graph.