0
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
What else can I help you with?
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 standard enthalpy of formation for N2 gas?
The standard enthalpy of formation for N2 gas is 0 kJ/mol.
What is the relationship between the standard enthalpy of formation and the standard enthalpy of reaction in chemical reactions?
The standard enthalpy of formation is the energy change when one mole of a compound is formed from its elements in their standard states. The standard enthalpy of reaction is the energy change for a reaction under standard conditions. The relationship between the two is that the standard enthalpy of reaction is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
How to calculate the enthalpy of a reaction?
To calculate the enthalpy of a reaction, you need to find the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. This is known as the enthalpy change (H) of the reaction. The enthalpy change can be determined using Hess's Law or by using standard enthalpy of formation values.
What is the standard enthalpy for sodium sulfate?
The standard enthalpy for sodium sulphate is -1387kJ/mol.
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 standard enthalpy of formation for N2 gas?
The standard enthalpy of formation for N2 gas is 0 kJ/mol.
What is the relationship between the standard enthalpy of formation and the standard enthalpy of reaction in chemical reactions?
The standard enthalpy of formation is the energy change when one mole of a compound is formed from its elements in their standard states. The standard enthalpy of reaction is the energy change for a reaction under standard conditions. The relationship between the two is that the standard enthalpy of reaction is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
How to calculate the enthalpy of a reaction?
To calculate the enthalpy of a reaction, you need to find the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. This is known as the enthalpy change (H) of the reaction. The enthalpy change can be determined using Hess's Law or by using standard enthalpy of formation values.
Why does oxygen gas not have an enthalpy of formation but ozone does?
Oxygen gas (O2) does not have an enthalpy of formation because it is an element in its standard state, which has an enthalpy of formation of zero by definition. Ozone (O3), on the other hand, is a compound and has a defined enthalpy of formation because it is formed from its elements in their standard states.
What is the standard enthalpy for sodium sulfate?
The standard enthalpy for sodium sulphate is -1387kJ/mol.
What is a standard formation reaction?
A standard formation reaction is the reaction of the ions which make up a compound in their standard states. For example, the standard formation reaction of H2SO4 is: 2H+(g) + SO42-(g) --> H2SO4 (l)
What is the process to calculate the delta H for the reaction?
To calculate the change in enthalpy (H) for a reaction, you need to subtract the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. This can be done using Hess's Law or by using standard enthalpy of formation values.
What is the standard enthalpy of formation of coconut oil?
The standard enthalpy of formation of coconut oil is not a well-defined value as it is a complex mixture of triglycerides, which are composed of various fatty acids. The enthalpy of formation of each individual fatty acid can be determined, but not for coconut oil as a whole.
What is the difference of enthalpy of formation and the enthalpy of combustion?
Enthalpy of combusion is energy change when reacting with oxygen. Enthalpy of formation is energy change when forming a compound. But some enthalpies can be equal.ex-Combusion of H2 and formation of H2O is equal
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
