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Reverse the equation for the decomposition given since you are doing formation. This makes -5678 positive.

Lookup the enthalpy of formation of gaseous water and carbon dioxide and multiply them by their moles in the formation equation. Remember nitrogen is in base form, so it is 0. You want:

Delta H of Reaction = Sum of Prod - Sum of Reactants

5678 = X - [-2418 + 0 + -4722 ]

X = -1462

-1462/4 = -365.5

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9y ago
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15y ago

DH= -5288.6 kJ/mol as per the CRC chem & physics handbook.

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15y ago

-5156.95kJ/mol

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11y ago

19.9 kJ/mo;e

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11y ago

20.3+-0.9 kj/mol

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10y ago

1.72 kJ/kg.K

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12y ago

76 kJ

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14y ago

76 kJ

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Q: Calculate the standard enthalpy of formation for nitroglycerin?
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Calculate standard enthalpy of formation of FeO?

You shouldn't "calculate" a standard enthalpy of formation. The beauty of standard enthalpies of formation is that they are already calculated for you. That is why they are delineated by the term "standard" - they are standards that were figured out by chemists some time ago, that never change, and can be found in tables usually in textbooks and even on Wikipedia. If you need to know the standard enthalpy of formation of FeO, Google it. And let me know what you find...because I can't seem to find a set answer either. I have found one site that lists the standard enthalpy of formation of FeO to be 271.9 kJ/mol. But it hasn't been so evident in other places. No wonder you were confused! Good luck.


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


What best describe the enthalpy of formation of a substance?

The enthalpy of formation of a substance is the amount of energy that was put in or evolved from the making of that substance from the individual elements.


How does lattice energy differs from bond enthalpy?

Bond enthalpy is the energy required to break a chemical bond whereas lattice energy is the enthalpy of formation of one mole of an ionic compound from gaseus ions under standard conditions.


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?

-1282.5J

Related questions

Calculate standard enthalpy of formation of FeO?

You shouldn't "calculate" a standard enthalpy of formation. The beauty of standard enthalpies of formation is that they are already calculated for you. That is why they are delineated by the term "standard" - they are standards that were figured out by chemists some time ago, that never change, and can be found in tables usually in textbooks and even on Wikipedia. If you need to know the standard enthalpy of formation of FeO, Google it. And let me know what you find...because I can't seem to find a set answer either. I have found one site that lists the standard enthalpy of formation of FeO to be 271.9 kJ/mol. But it hasn't been so evident in other places. No wonder you were confused! Good luck.


What is the heat of formation of SO2?

The standard heat/enthalpy of formation of SO2 is -296.8 KJ


WHAT IS THE STANDARD HEAT FORMATION OF POTASSIUM HYDROXIDE (S)?

The standard enthalpy of formation for potassium hydroxide is -425,8 kJ/mol.


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.


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


Why is an input of energy needed when forming NaCl?

The standard enthalpy of formation for NaCl solid is: -411,12 kJ/mol at 25 0C.


What is the standard molar enthalpy of formation for ammonium chloride aq?

-299.65 kJ/mole


What best describe the enthalpy of formation of a substance?

The enthalpy of formation of a substance is the amount of energy that was put in or evolved from the making of that substance from the individual elements.


What is the standard enthalpy of formation of NaF when 56 g of Na reacts with excess F?

6


What are three termodynamic properties?

- Gibbs energy- Standard enthalpy of formation- Specific heat capacity


Which has a lower enthalpy ethylen oxide or acetaldehyde?

The standard enthapy of formation for ethylene oxide is -52,6 J/mol.The standard enthapy of formation for acetaldehyde is is -166,6 J/mol.


What is the standard enthalpy of formation of hydrogen bromide?

[from wikipedia] The standard enthalpy of formation"standard heat of formation" of a compound is the change of enthalpy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at 1 bar of pressure and the specified temperature, usually 298.15 K or 25 degrees Celsius). Its symbol is ΔHfO.