Consider the reaction 2aloh3sal2o3s 3h2ol with enthalpy of reaction hrxn210kjmol


What is the reaction?


2Al(OH)3 + H2SO4 -> Al2(SO4)3 + 3H2O

This is a double replacement reaction in which Al(OH)3 (aluminium hydroxide) and H2SO4 (sulfuric acid) react to form Al2(SO4)3 (aluminium sulfate) and H2O (water).

What is the enthalpy of the reaction?


In order to calculate the enthalpy of the reaction, we need to know the enthalpies of formation of all of the products and reactants. The enthalpy of formation is the heat required to form one mole of a substance from its elements in their standard states.

2AlOH3(s) + 3H2O(l) —> Al2O3(s) + 6H2O(l)

The standard enthalpy of formation for Al2O3(s) is -1675 kJ/mol and for H2O(l) it is -285.8 kJ/mol.

Therefore, the enthalpy of the reaction is:

-1675 kJ/mol + (6 x -285.8 kJ/mol) = -210 kJ/mol

What is the standard enthalpy of formation for the products?


In order to find the standard enthalpy of formation for the products, we need to find the change in enthalpy for the reaction. We can do this by using the given information and the equation:

ΔHrxn=ΔHf(products)-ΔHf(reactants)

We can plug in the given information to solve for the change in enthalpy of formation for the products.

ΔHrxn=210 kJ/mol
ΔHf(reactants)=0 kJ/mol

This leaves us with ΔHf(products)=210 kJ/mol.

What is the standard enthalpy of formation for the reactants?

In order to calculate the standard enthalpy of formation for the reactants, we must first determine the molar enthalpy of each reactant. For reactant A, we have 2 mol Al(OH)3 x 105.5 kJ/mol = 211 kJ. For reactant B, we have 3 mol Na2SO3 x 109.3 kJ/mol = 327.9 kJ. Therefore, the standard enthalpy of formation for the reactants is 538.9 kJ/mol.

What is the heat of the reaction?

We can calculate the heat of the reaction using Hess’s Law. In this case, we know the enthalpy of formation for each reactant and product.

The heat of reaction is calculated as:

heat of reaction = [(enthalpy of products) – (enthalpy of reactants)] x (stoichiometric coefficient)

In this case, the stoichiometric coefficients are all 1. So, we have:

heat of reaction = [(enthalpy of 3H2O(l)) + (enthalpy of 2Al(OH)3(s)) + (enthalpy of 3Sal2O3(s))] – [(enthalpy of 2Al(OH)3(s)) + (enthalpy of 3Sal2O3(s)) + (enthalpy of 3H2O(l))]

heat of reaction = 0 kJ

What is the standard heat of formation for the products?


To calculate the standard heat of formation for the products, we need to know the standard enthalpies of formation for all of the reactants and products. The standard enthalpy of formation is the change in enthalpy that occurs when one mole of a substance is formed from its element in its standard state.

The products of this reaction are 2Al(OH)3, 3H2O and 2NaClO3. The standard enthalpies of formation for these substances can be found in a table of thermodynamic data.

From this information, we can calculate the standard heat of formation for the products as follows:

ΔHrxn = ΔHf(products) – ΔHf(reactants)

ΔHrxn = [2 x -466.8 kJ/mol] + [3 x -285.8 kJ/mol] + [2 x -1117 kJ/mol] – [210 kJ/mol]

ΔHrxn = -1671.4 kJ/mol

What is the standard heat of formation for the reactants?


In order to calculate the standard heat of formation for the reactants, we need to know the standard enthalpies of formation for each reactant. The standard enthalpy of formation is the heat change that occurs when one mole of a compound is formed from its elements in their standard states.

The standard enthalpy of formation for water is -285.8 kJ/mol and the standard enthalpy of formation for aluminum hydroxide is -1669 kJ/mol. This means that the standard heat of formation for the reactants is -954.8 kJ/mol.

How do you calculate the heat of the reaction?


In order to calculate the heat of the reaction, you need to know the enthalpy of each reactant and product. The enthalpy change of the reaction can be calculated using the following equation:

ΔHrxn = ΣΔHf products – ΣΔHf reactants

where ΔHf is the standard enthalpy of formation.

What is the molar heat of the reaction?

The molar heat of the reaction is calculated by subtracting the enthalpy of the products from the enthalpy of the reactants. In this case, the enthalpy of the products is -210 kJ/mol, and the enthalpy of the reactants is 0 kJ/mol. Therefore, the molar heat of the reaction is 210 kJ/mol.

What is the standard molar heat of formation for the products?


In order to calculate the standard molar heat of formation for the products, we need to know the enthalpy of reaction and the number of moles of each product.

The enthalpy of reaction is given as 210 kJ/mol. This means that for every 1 mole of reactants, 210 kJ of energy is released.

The products are aluminum oxide, water, and sulfuric acid. We can assume that there are 2 moles of aluminum oxide, 3 moles of water, and 1 mole of sulfuric acid.

Aluminum oxide has a standard molar heat of formation of -1673 kJ/mol. Water has a standard molar heat of formation of -285.8 kJ/mol. Sulfuric acid has a standard molar heat of formation of -890 kJ/mol.

When we plug these values into the equation, we get:

-1673 kJ/mol x 2 + (-285.8 kJ/mol x 3) + (-890 kJ/mol) = -3210.4 kJ/mol

What is the standard molar heat of formation for the reactants?

The standard molar heat of formation for the reactants can be calculated using the following equation:

HF(reactants) = HRXN (210 kJ/mol) / (2 mol AlOH3 + 3 mol Na2O3S + 3 mol H2O)

HF(reactants) = 210 kJ/mol / 8 mol

HF(reactants) = 26.25 kJ/mol

How do you calculate the molar heat of the reaction?

You can calculate the molar heat of the reaction by looking at the enthalpy of reaction. The enthalpy of reaction (ΔHrxn) is the heat change that occurs when one mole of a substance reacts. To calculate the molar heat of the reaction, you would need to know the enthalpy of the reactants and products, as well as the stoichiometry of the reaction.


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