Calculate the standard enthalpy change for the following reaction at 25 °C.
MgCl2(s) + H2o(l) ---> MgO(s) + 2HCl(g)
For anyone who stumbles upon this page, DrBob222 made a mistake. Take note that the second reactant is H2O (l), not another 2HCl (g). Make sure to look up the dHf of H2O instead.
Could you elaborate? I don't quite understand what that means.
MgCl2(s) + H2o(l) ---> MgO(s) + 2HCl(g)
Look up delta H formation for MgO(solid), HC(gas), MgCl2(solid), and H2O(liquid).
Then n = mols and those are the coefficients. The products are MgCl2 and HCl. Reactants are MgCl2 and H2O.
So you have (1*dHf MgO + 2*dHf HCl)as the products and subtract from that (1*dHf MgCl2 + 2*dHf HCl) as the reactants. That will give you the dH reaction. It all goes together as this.
dHrxn = (n*dHf products) - (n*dHf reactants)
Well, I haven't seen this many letters since I got a ransom note from my pet parrot. But don't worry, I'll crunch some numbers for you!
To find the standard enthalpy change (ΔH°) for the reaction, we need to look up the standard enthalpies of formation (∆H°f) for the reactants and products. The standard enthalpy change is the difference between the sum of the products' enthalpies of formation and the sum of the reactants' enthalpies of formation.
So, let's find the standard enthalpies of formation for each compound involved:
ΔH°f(MgCl2(s)) = -641 kJ/mol
ΔH°f(H2O(l)) = -286 kJ/mol
ΔH°f(MgO(s)) = -601 kJ/mol
ΔH°f(HCl(g)) = -92.3 kJ/mol
Now let's calculate the ΔH° for the reaction:
Reactants = ΔH°f(MgCl2(s)) + ΔH°f(H2O(l))
Products = ΔH°f(MgO(s)) + 2 * ΔH°f(HCl(g))
ΔH° = Products - Reactants
ΔH° = [ΔH°f(MgO(s)) + 2 * ΔH°f(HCl(g))] - [ΔH°f(MgCl2(s)) + ΔH°f(H2O(l))]
ΔH° = [-601 kJ/mol + 2(-92.3 kJ/mol)] - [-641 kJ/mol + (-286 kJ/mol)]
Calculating that mess gives me a value of ΔH° ≈ -461.4 kJ/mol. That's the standard enthalpy change for the reaction at 25 °C.
Remember, this is a clownbot calculating things, so take it with a pinch of laughter!
To calculate the standard enthalpy change for the given reaction at 25 °C, you would need to know the standard enthalpy of formation values for each of the substances involved in the reaction. The standard enthalpy of formation (ΔHf°) is the change in enthalpy that occurs when one mole of a substance is formed from its elements in their standard states, with all reactants and products at a pressure of 1 bar.
The standard enthalpy change (ΔH°) for a reaction can be calculated using Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the system.
To calculate the standard enthalpy change for the given reaction, you would follow these steps:
1. Identify the standard enthalpy of formation values.
- ΔHf°(MgCl2): Standard enthalpy of formation of MgCl2
- ΔHf°(H2O): Standard enthalpy of formation of H2O
- ΔHf°(MgO): Standard enthalpy of formation of MgO
- ΔHf°(HCl): Standard enthalpy of formation of HCl
2. Determine the balanced chemical equation for the reaction.
MgCl2(s) + H2O(l) → MgO(s) + 2HCl(g)
3. Apply Hess's Law.
The standard enthalpy change for the reaction can be calculated using the equation:
ΔHrxn° = ΣnΔHf°(products) - ΣnΔHf°(reactants)
where Σn is the sum of the stoichiometric coefficients of the products or reactants.
4. Substitute the values of standard enthalpy of formation into the equation and perform the calculation to obtain the standard enthalpy change of the reaction at 25 °C.
Note: The standard enthalpy of formation values are usually provided in reference tables or databases.