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Calculate the change in enthalpy for the Mg HCl reaction using standard enthalpies of formation delta hf for HCl aq equals -167.2 kj and delta hf for MgCl2 aq equals -791.2 kj?
Clark21
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∙ 10y ago
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How is the enthalpy of reaction related to the enthalpies of formation for the products and reactants?
The difference between the enthalpy of formation of the products minus the enthalpy of formation of the reactants is the enthalpy of the reaction
How bond formation and bond breaking related to enthalpy?
Enthalpy is the measurement of total energy change of a reaction. The energy of bond formation and bond breaking can be used to calculate the bond enthalpy of the reaction. Bond enthalpy is the enthalphy change when 1 mol of bond is broken. Therefore the general equation to calculate the enthalpy change is energy of bond broken subtract by energy of bond formation.
How is Hess's law used to measures enthalpy of a desired reaction?
... Intermediate equations with known enthalpies are added together.
How is Hess's law used to measure enthalpy of desired reaction?
Intermediate equations with known enthalpies are added together
How is Hess's law used to measure enthalpy of a desired reaction?
... Intermediate equations with known enthalpies are added together.
How is the enthalpy of reaction related to the enthalpies of formation for the products and reactants?
The difference between the enthalpy of formation of the products minus the enthalpy of formation of the reactants is the enthalpy of the reaction
What does Hesses law state?
The enthalpy of a reaction is the sum of the enthalpies of intermediate reaction.
How bond formation and bond breaking related to enthalpy?
Enthalpy is the measurement of total energy change of a reaction. The energy of bond formation and bond breaking can be used to calculate the bond enthalpy of the reaction. Bond enthalpy is the enthalphy change when 1 mol of bond is broken. Therefore the general equation to calculate the enthalpy change is energy of bond broken subtract by energy of bond formation.
What is the purpose of defining the standard enthalpies of formation?
The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements. A triangle is a change in enthalpy. A degree signifies that it's a standard enthalpy change. A f is a reaction from a substance that's formed from its elements.
How is Hess's law used to measure enthalpy of a desired reaction?
... Intermediate equations with known enthalpies are added together.
How is Hess's law used to measure enthalpy of a desire reaction?
... Intermediate equations with known enthalpies are added together.
How is hess law used to measure enthalpy of a desired reaction?
... Intermediate equations with known enthalpies are added together.
How is Hess's law used to measures enthalpy of a desired reaction?
... Intermediate equations with known enthalpies are added together.
How is Hess's law used to measure enthalpy of desired reaction?
Intermediate equations with known enthalpies are added together
What is the basic difference between enthalpy of formation?
Standard Heat (Enthalpy) of Formation, Hfo, of any compound is the enthalpy change of the reaction by which it is formed from its elements, reactants and products all being in a given standard state.By definition, the standard enthalpy (heat) of formation of an element in its standard state is zero, Hfo = 0.Standard Molar Enthalpy (Heat) of Formation, Hmo, of a compound is the enthalpy change that occurs when one mole of the compound in its standard state is formed from its elements in their standard states.Standard Enthalpy (Heat) of Reaction, Ho, is the difference between the standard enthalpies (heats) of formation of the products and the reactants.Ho(reaction) = the sum of the enthalpy (heat) of formation of products - the sum of the enthalpy (heat) of formation of reactants: Ho(reaction) = Hof(products) - Hof(reactants)To calculate an Enthalpy (Heat) of Reaction:Write the balanced chemical equation for the reaction Remember to include the state (solid, liquid, gas, or aqueous) for each reactant and product.Write the general equation for calculating the enthalpy (heat) of reaction: Ho(reaction) = Hof(products) - Hof(reactants)Substitute the values for the enthalpy (heat) of formation of each product and reactant into the equation. Remember, if there are 2 moles of a reactant or product, you will need to multiply the enthalpy term by 2, if molar enthalpies (heats) of formation are used.Standard Enthalpy (Heat) of FormationExample: Standard Enthalpy (Heat) of Formation of WaterThe standard enthalpy (heat) of formation for liquid water at 298K (25o) is -286 kJ mol-1. This means that 286 kJ of energy is released when liquid water, H2O(l), is produced from its elements, hydrogen and oxygen, in their standard states, ie, H2(g) and O2(g).This reaction is written as:H2(g) + ½O2(g) -----> H2O(l) Hfo = -286 kJ mol-1The standard enthalpy (heat) of formation of water vapour at 298K (25o) is -242 kJ mol-1.This means that 242 kJ of energy is released when gaseous water (water vapour), H2O(g), is produced from its elements, hydrogen and oxygen, in their standard states, ie, H2(g) and O2(g).This reaction is written as:H2(g) + ½O2(g) -----> H2O(g) Hfo = -242 kJ mol-1
Which equation do you use to calculate the enthalpy of neutralisation from an acid-base reaction?
yea
Would this reaction happen NaCl plus CaF2 - NaF plus CaCl2?
The reaction is2NaCl + CaF2 --> 2NaF + CaCl2and the enthalpies of formation (kJ/Mol) of the solids at STP are2NaCl = (-411)*2CaF2 = -1220Total = -20422NaF = (-469)*2CaCl2 = -796Total = -1734The calculations show that the reactants are more stable than the products by some 307kJ/mol, so that the reaction would not proceed. (For more advanced readers, the value for the Gibbs free energy of formation will not be very much different from this, as the entropy terms will be relatively small in comparison with the enthalpy terms.) A major factor in this is the relatively high lattice enthalpy of CaF2.
