Given that:
2Al(s)+(3/2)O2(g)--->Al2O3(s)
change in H(rxn)= -1601 kJ/mol
2Fe(s)+(3/2)O2(g)--->Fe2O3(s)
change in H(rxn)= -821 kJ/mol
Calculate the standard enthalpy change for the following reaction:
2Al(s)+Fe2O3(s)--->2Fe(s)+Al2O3(s)
_______kJ
PLEASE EXPLAIN IM SOOO CONFUSED, CAN YOU SHOW ME THE STEPS?? Thank you
Sure, I can help you with that! To calculate the standard enthalpy change for the given reaction, you can apply the Hess's Law equation:
ΔH(rxn) = Σ(ΔH(products)) - Σ(ΔH(reactants))
In simpler terms, you need to find the sum of the enthalpy changes of the products and subtract the sum of the enthalpy changes of the reactants.
Let's start by writing the given equations and their enthalpy changes again for reference:
Equation 1: 2Al(s) + (3/2)O2(g) ---> Al2O3(s) (ΔH1 = -1601 kJ/mol)
Equation 2: 2Fe(s) + (3/2)O2(g) ---> Fe2O3(s) (ΔH2 = -821 kJ/mol)
Now, we need to manipulate these equations in order to obtain our target equation:
Equation 1 multiplied by 2: 4Al(s) + 3O2(g) ---> 2Al2O3(s) (Multiply ΔH1 by 2)
Equation 2 multiplied by 2: 4Fe(s) + 3O2(g) ---> 2Fe2O3(s) (Multiply ΔH2 by 2)
By manipulating these equations, we can see that we can cancel out the O2(g) term. Now, let's combine these new equations:
4Al(s) + 3O2(g) ---> 2Al2O3(s) (ΔH1 = -3202 kJ/mol)
4Fe(s) + 3O2(g) ---> 2Fe2O3(s) (ΔH2 = -1642 kJ/mol)
Now let's write the target equation:
2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s)
Since we have 2 moles of Al2O3(s) and Fe2O3(s) in the products, we need to multiply the ΔH2 by 2. Likewise, since we have 2 moles of Al(s) and Fe(s) in the reactants, we need to multiply the ΔH1 by 2. The O2(g) term is cancelled out, so it doesn't contribute to the overall enthalpy change.
Finally, let's calculate the standard enthalpy change using the Hess's Law equation:
ΔH(rxn) = (2 * ΔH1) + (2 * ΔH2)
= (2 * -3202 kJ/mol) + (2 * -1642 kJ/mol)
= -6404 kJ/mol - 3284 kJ/mol
= -9688 kJ/mol
Therefore, the standard enthalpy change for the given reaction 2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s) is -9688 kJ/mol.
I hope this explanation clarifies your confusion! Let me know if you have any other questions.
To calculate the standard enthalpy change for the reaction 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3(s), we can use the concept of Hess's Law, which states that the total enthalpy change of a reaction is independent of the pathway taken.
1. Write down the given reactions and their corresponding enthalpy changes:
Reaction 1: 2Al(s) + (3/2)O2(g) -> Al2O3(s) (enthalpy change = -1601 kJ/mol)
Reaction 2: 2Fe(s) + (3/2)O2(g) -> Fe2O3(s) (enthalpy change = -821 kJ/mol)
2. Reverse Reaction 1 and multiply it by 2 to match the coefficients of Fe2O3:
-2Al2O3(s) -> 4Al(s) + 3O2(g) (enthalpy change = 2 * 1601 kJ/mol = -3202 kJ/mol)
3. Add Reaction 2 and the reversed Reaction 1 to get the overall reaction:
2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3(s) (enthalpy change = -821 kJ/mol - 3202 kJ/mol = -4023 kJ/mol)
Therefore, the standard enthalpy change for the reaction 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3(s) is -4023 kJ.
To calculate the standard enthalpy change for the given reaction, we can use Hess's Law. Hess's Law states that the overall change in enthalpy for a reaction is equal to the sum of the individual enthalpy changes for each step of the reaction, regardless of the path taken.
We can break down the given reaction into two steps:
Step 1: Al(s) + (3/2)O2(g) ---> Al2O3(s)
Step 2: 2Fe(s) + Al2O3(s) ---> Fe2O3(s) + Al2O3(s)
Now, we need to use the given enthalpy changes for each step:
For Step 1:
2Al(s) + (3/2)O2(g) ---> Al2O3(s) (given change in H(rxn) = -1601 kJ/mol)
For Step 2:
2Fe(s) + Al2O3(s) ---> Fe2O3(s) + Al2O3(s) (given change in H(rxn) = -821 kJ/mol)
To calculate the enthalpy change for the overall reaction, we need to manipulate the given equations so that they match the overall reaction. We'll multiply the first equation by 2 to balance the number of Al atoms, and multiply the second equation by 3 to balance the number of Al2O3 formed:
Step 1: 2Al(s) + 3O2(g) ---> 2Al2O3(s)
Step 2: 6Fe(s) + 2Al2O3(s) ---> 3Fe2O3(s) + 2Al2O3(s)
Now, we can add the two equations together, canceling out the common reactant and product:
2Al(s) + 3O2(g) + 6Fe(s) + 2Al2O3(s) ---> 2Al2O3(s) + 3Fe2O3(s) + 2Al2O3(s)
Simplifying this equation, we get:
2Al(s) + 3O2(g) + 6Fe(s) ---> 3Fe2O3(s) + 4Al2O3(s)
Now, we can calculate the enthalpy change for the overall reaction:
Change in H(rxn) = 2 * (-1601 kJ/mol) + 3 * (-821 kJ/mol)
= -3202 kJ/mol - 2463 kJ/mol
= -5665 kJ/mol
Therefore, the standard enthalpy change for the reaction 2Al(s) + Fe2O3(s) ---> 2Fe(s) + Al2O3(s) is -5665 kJ/mol.