A considerable amount of heat is required for the decomposition of aluminum oxide shown below. What is the heat change when 3.5 moles of aluminum oxide are decomposed? Is this endothermic or exothermic?
2Al2O3(s) → 4Al(s) + 3O2(g) ΔH = 3352kJ
2Al2O3(s) → 4Al(s) + 3O2(g) ΔH = 3352kJ
If the reaction shown is 3352 kJ FOR THE REACTION AS SHOWN, then
dH for the question = 3352 kJ x (3.5/2) = ?
If dH is + this is an endothermic reaction.
To find the heat change when 3.5 moles of aluminum oxide are decomposed, we need to use the given ΔH value.
Given: ΔH = 3352 kJ
The balanced equation shows that 2 moles of aluminum oxide produce 3352 kJ of heat. Therefore, we can set up a proportion to find the heat change for 3.5 moles.
2 mol Al2O3 = 3352 kJ
3.5 mol Al2O3 = x (kJ)
Using the proportion:
3.5 mol Al2O3 / 2 mol Al2O3 = x (kJ) / 3352 kJ
Cross multiplying, we get:
3.5 mol Al2O3 * 3352 kJ = 2 mol Al2O3 * x (kJ)
x = (3.5 mol Al2O3 * 3352 kJ) / 2 mol Al2O3
x ≈ 5866 kJ
Therefore, the heat change when 3.5 moles of aluminum oxide are decomposed is approximately 5866 kJ.
To determine if this reaction is endothermic or exothermic, we can look at the sign of ΔH. In this case, the given ΔH value is positive (3352 kJ), indicating that the reaction is endothermic.
To calculate the heat change (ΔH) when 3.5 moles of aluminum oxide are decomposed, you can use the equation:
Heat change = (ΔH / moles of reaction) x moles of substance
Given that the ΔH for the reaction is 3352 kJ, and the reaction involves the decomposition of 2 moles of aluminum oxide, we can calculate:
Heat change = (3352 kJ / 2 mol) x 3.5 mol
Heat change = 5028 kJ
Therefore, the heat change when 3.5 moles of aluminum oxide are decomposed is 5028 kJ.
To determine whether the reaction is endothermic or exothermic, we can look at the sign of the heat change. If the heat change is positive, the reaction is endothermic (heat is absorbed). If the heat change is negative, the reaction is exothermic (heat is released).
In this case, since the heat change is positive (5028 kJ), the reaction is endothermic.