the standard enthalpy change of vaporization DHov is the enthalpy change when one mole of a substance is transformed into a gas enthalpy change is the term we use to describe the energy exchange that occurs with the surroundings at a constant temperature and pressure so to work it out, use the formula DH = cmDT DH - the enthalpy change c - the specific heat capacity of butanol (kJ kg-1 °C-1) m - the mass of butanol heated (kg) DT - the change in temperature of the butanol (°C)
so there is no general enthalpy change of butanol, it depends on the factors above. the specific heat capacity of butanol, the mass of butanol heated, and the change in temperature of the butanol should be given to you in order to work the enthalpy change of vaporization of butanol if there is a rise in temperature, the reaction is exothermic and if there is a drop in temperature the reaction is endothermic. exothermic reactions have a negative enthalpy change, and therefore endothermic reactions have a positive enthalpy change. hope it helped (:
The standard enthalpy of formation for N2 gas is 0 kJ/mol.
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.
n-butane: -140.7 kJ/mol (liq.) & -124.7 kJ/mol (gas)isobutane: -158.4 kJ/mol (liq.) & -134.5 kJ/mol (gas)
The standard enthalpy for sodium sulphate is -1387kJ/mol.
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
C4H10(g) is about -126 kJ/mol C4H10(l) is about -147 kJ/mol
The standard enthalpy of formation for N2 gas is 0 kJ/mol.
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.
n-butane: -140.7 kJ/mol (liq.) & -124.7 kJ/mol (gas)isobutane: -158.4 kJ/mol (liq.) & -134.5 kJ/mol (gas)
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.
The standard enthalpy for sodium sulphate is -1387kJ/mol.
To calculate the enthalpy of formation of Cl2NNF2(g), you can use the standard enthalpy of formation values of the reactants and products involved in the relevant chemical reaction. The enthalpy of formation is determined by the equation: ΔH_f° = ΣΔH_f°(products) - ΣΔH_f°(reactants). You need to find the standard enthalpy of formation for Cl2NNF2(g) and the standard enthalpies of the elements in their standard states (Cl2, N2, and F2) to perform this calculation. If the standard enthalpy values are not available, you may need to use Hess's law and related reactions to derive the value indirectly.
The enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. It is a measure of the energy released or absorbed during the formation of a substance. This value is crucial in thermodynamics and helps in predicting the feasibility of chemical reactions. The standard enthalpy of formation for elements in their standard states is defined as zero.
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.
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
To calculate the enthalpy change of formation from combustion, you can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps. First, determine the enthalpy change for the combustion reaction using a calorimeter or from standard enthalpy values. Then, apply the equation: ΔH_f = ΔH_combustion + Σ(ΔH_f of products) - Σ(ΔH_f of reactants), where ΔH_f is the standard enthalpy of formation. This allows you to derive the enthalpy of formation for the desired compound based on its combustion data.
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.