Calculate the heat of decomposition for this process at constant pressure and 25°C. (Look up the standard enthalpy of formation of the reactant and products in the table; ΔH°f(BaCO3(s)) = -1218.8 kJ/mol, ΔH°f (BaO(s)) = -558.2 kJ/mol and ΔH°f (Na2CO3(s)) = -1130.9 kJ/mol, ΔH°f (Na2O(s)) = -415.89 kJ/mol.)

MgCO3(s) → MgO(s) + CO2(g)

Well, it seems like we have a chemical reaction going on here! Let's calculate the heat of decomposition, shall we?

First, we need to determine the enthalpy change for the reaction by subtracting the sum of the enthalpies of formation of the products from the sum of the enthalpies of formation of the reactant.

ΔH° = [ΔH°f (MgO(s))] + [ΔH°f (CO2(g))] - [ΔH°f (MgCO3(s))]

Now, let's look up the values in the table and plug them in:

ΔH° = [(-558.2 kJ/mol) + (-415.89 kJ/mol)] - [(-1218.8 kJ/mol)]

ΔH° = -973.09 kJ/mol

So, the heat of decomposition for this process at constant pressure and 25°C is approximately -973.09 kJ/mol. That's quite a big release of heat! Maybe that's why things can get hot and steamy in chemistry sometimes!

To calculate the heat of decomposition for the given process, we need to use the standard enthalpy of formation values for the reactant and products.

The balanced chemical equation for the reaction is:
MgCO3(s) → MgO(s) + CO2(g)

The enthalpy change for this reaction can be calculated using the equation:

ΔH° = ΣΔH°f (products) - ΣΔH°f (reactants)

Let's substitute the values we were given:

Reactant:
ΔH°f (MgCO3(s)) = 0 kJ/mol (since it is not specified in the table)

Products:
ΔH°f (MgO(s)) = -558.2 kJ/mol
ΔH°f (CO2(g)) = 0 kJ/mol (since it is a gas, usually considered as the standard state)

Now we can calculate the heat of decomposition:

ΔH° = ΣΔH°f (products) - ΣΔH°f (reactants)
= [ΔH°f (MgO(s)) + ΔH°f (CO2(g))] - [ΔH°f (MgCO3(s))]
= [-558.2 kJ/mol + 0 kJ/mol] - [0 kJ/mol]
= -558.2 kJ/mol

Therefore, the heat of decomposition for the given process at constant pressure and 25°C is -558.2 kJ/mol.

To calculate the heat of decomposition for the given process, we need to use the standard enthalpy of formation values for each compound involved. The standard enthalpy of formation, denoted as ΔH°f, is the heat change that occurs when one mole of a compound is formed from its elements in their standard states.

The balanced equation for the decomposition of MgCO3 is:
MgCO3(s) → MgO(s) + CO2(g)

Now we can calculate the heat of decomposition using the enthalpy of formation values for the reactant and products:

ΔH°decomposition = Σ(ΔH°f(products)) - ΔH°f(reactant)

First, we need the enthalpy of formation values for each compound involved in the reaction:

ΔH°f(MgCO3(s)) = ?
ΔH°f(MgO(s)) = -558.2 kJ/mol
ΔH°f(CO2(g)) = ?

We can find the missing enthalpy of formation value for MgCO3 by subtracting the enthalpy of formation values for MgO and CO2 from the given enthalpy of formation value for BaCO3:

ΔH°f(MgCO3(s)) = ΔH°f(BaCO3(s)) - ΔH°f(BaO(s)) - ΔH°f(Na2CO3(s))

Substituting the given values:
ΔH°f(MgCO3(s)) = -1218.8 kJ/mol - (-558.2 kJ/mol) - (-1130.9 kJ/mol)
= -1218.8 kJ/mol + 558.2 kJ/mol - 1130.9 kJ/mol
= -1791.7 kJ/mol

Now that we have all the enthalpy of formation values, we can calculate the heat of decomposition:

ΔH°decomposition = ΔH°f(MgO(s)) + ΔH°f(CO2(g)) - ΔH°f(MgCO3(s))
= -558.2 kJ/mol + ΔH°f(CO2(g)) - (-1791.7 kJ/mol)

To find the enthalpy of formation for CO2, we can use the enthalpy of formation values for Na2CO3 and Na2O and rearrange the equation:

ΔH°f(CO2(g)) = ΔH°f(Na2CO3(s)) - 2 * ΔH°f(Na2O(s))

Substituting the given values:
ΔH°f(CO2(g)) = -1130.9 kJ/mol - 2 * (-415.89 kJ/mol)
= -1130.9 kJ/mol + 831.78 kJ/mol
= -299.12 kJ/mol

Now we can substitute this value back into the heat of decomposition equation:

ΔH°decomposition = -558.2 kJ/mol + (-299.12 kJ/mol) - (-1791.7 kJ/mol)

Simplifying further:
ΔH°decomposition = -558.2 kJ/mol + 299.12 kJ/mol + 1791.7 kJ/mol
= 1532.62 kJ/mol

Therefore, the heat of decomposition for the given process at constant pressure and 25°C is 1532.62 kJ/mol.