Suppose we had mixed 52.37 mL of 1.00 M HI solution with 25.42 mL of 1.00 M KOH solution, and measured a heat of reaction of -1.52 kJ. What is ΔH for this reaction, in kJ? Assume that all solutions have the same density as water (1.00 g/mL).
HI + KOH ==>KI + H2O
mols HI = M x L = ?
mols KOH = M x L = ?
One of them will be the limiting reagent; i.e., the one with the smaller mols. That will give you the mols of H2O produced.
You know q is 1.52 kJ and that is dHrxn. USUALLY you want kJ/mol. Then dHrxn/mols H2O will give that to you.
To find the enthalpy change (ΔH) for the reaction, we can use the equation:
ΔH = q / n
Where:
q = heat of reaction
n = number of moles of the limiting reactant
First, we need to determine the limiting reactant. To do this, we can compare the number of moles of HI and KOH in each solution.
For HI solution:
Volume = 52.37 mL
Density = 1.00 g/mL
Mass of HI = Volume × Density
= 52.37 mL × 1.00 g/mL
= 52.37 g
Molar mass of HI = 127.91 g/mol
Number of moles of HI = Mass / Molar mass
= 52.37 g / 127.91 g/mol
= 0.409 mol
For KOH solution:
Volume = 25.42 mL
Density = 1.00 g/mL
Mass of KOH = Volume × Density
= 25.42 mL × 1.00 g/mL
= 25.42 g
Molar mass of KOH = 56.11 g/mol
Number of moles of KOH = Mass / Molar mass
= 25.42 g / 56.11 g/mol
= 0.454 mol
Since HI and KOH have a 1:1 stoichiometry, the number of moles of HI is less than the number of moles of KOH. Therefore, HI is the limiting reactant.
Now we can calculate the enthalpy change (ΔH) using the given heat of reaction.
ΔH = q / n
= -1.52 kJ / 0.409 mol
= -3.72 kJ/mol
Thus, the enthalpy change (ΔH) for this reaction is -3.72 kJ/mol.
To find the change in enthalpy (ΔH) for the reaction, we can use the equation:
ΔH = q / (moles of limiting reactant)
First, we need to determine the limiting reactant. This is the reactant that will be consumed completely and limits the amount of product that can be formed. To do this, we can calculate the number of moles of each reactant.
Let's start by finding the moles of HI solution:
Moles of HI = Volume of HI solution (in L) x Molarity of HI solution
= 52.37 mL / 1000 mL/L x 1.00 mol/L
= 0.05237 mol
Next, let's calculate the moles of KOH solution:
Moles of KOH = Volume of KOH solution (in L) x Molarity of KOH solution
= 25.42 mL / 1000 mL/L x 1.00 mol/L
= 0.02542 mol
Based on the balanced equation, the molar ratio between HI and KOH is 1:1. Therefore, the number of moles of HI and KOH are the same.
Since the moles of HI and KOH are equal, the limiting reactant is the one with the smaller initial moles, which is KOH.
Now, let's calculate the heat of reaction (q). The given value, -1.52 kJ, represents the heat released by the reaction.
Next, we can substitute the values into the equation to find ΔH:
ΔH = q / (moles of limiting reactant)
= -1.52 kJ / 0.02542 mol
≈ -59.74 kJ
Therefore, the ΔH for this reaction is approximately -59.74 kJ.