-486kj
Enthalpy is the energy absorbed or lost from a reaction, but enthalpy change per mole is the amount of energy lost per mole, so in order to get the overall enthalpy from the change per mole, you must multiply that value by the amount of moles used in the reaction.
You need to know the ∆H of this reaction, or the enthalpy change. If you know the enthalpy change, then convert the 6.44 g of sulfur to moles, and use stoichiometry to determine what fraction of a mole was reacted, and multiply that time the value of ∆H of the reaction
The answer to this question will depend on what the substance that is reacting is. You will need to find the appropriate standard enthalpy value, which corresponds to the amount of enthalpy change when one mole of matter is transformed by a chemical reaction in standard conditions.
The thermal decomposition reaction is:2 KO2------------K2O2 + O20,2 moles of O2 are produced from o,4 moles KO2.
The complete decomposition reaction is as follows:2 BrF3 → Br2 + 3 F2 , so 2 moles BrF3 will give 1 mole Br2 , hence 0.248 mole gives 0.124 mole Br2
Enthalpy is the energy absorbed or lost from a reaction, but enthalpy change per mole is the amount of energy lost per mole, so in order to get the overall enthalpy from the change per mole, you must multiply that value by the amount of moles used in the reaction.
You need to know the ∆H of this reaction, or the enthalpy change. If you know the enthalpy change, then convert the 6.44 g of sulfur to moles, and use stoichiometry to determine what fraction of a mole was reacted, and multiply that time the value of ∆H of the reaction
This enthalpy of reaction is 560 joules, for liquid water at 25 oC.
The answer to this question will depend on what the substance that is reacting is. You will need to find the appropriate standard enthalpy value, which corresponds to the amount of enthalpy change when one mole of matter is transformed by a chemical reaction in standard conditions.
The thermal decomposition reaction is:2 KO2------------K2O2 + O20,2 moles of O2 are produced from o,4 moles KO2.
Reactions that increase the randomness. Reactions that have more moles of gas on the product side than the reactant side increase entropy. Also reactions that have a positive change in spontaneity and a negative enthalpy.
The complete decomposition reaction is as follows:2 BrF3 → Br2 + 3 F2 , so 2 moles BrF3 will give 1 mole Br2 , hence 0.248 mole gives 0.124 mole Br2
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
The total number of moles of gas on each side of the reaction.
Assuming a decomposition reaction with this equation: 2KClO3(s) --> 2KCl(s) + 3O2(g), the ratio is 2:3, and if you produce 15mol O2, then 10mol potassium chlorate are needed.
sulphuric acid is a diprotic acid so has two H+ and needs two moles of sulphuric acid to neutralise it
O2 is an oxygen molecule, H2O is a water molecule. The 2 in front of each means that you have 2 moles of each