A 65.0 g sample of 0.600 M HI at 18.46 degrees celsius is added to 84.0 g of solution containing excess KOH, also at 18.46 degrees celsius. The final temperature is 21.96 degrees celsius. Calculate the enthalpy of H for the reaction.
Assume the specific heat of the solution is 4.184 J/g(degrees celsius), and the calorimeter has negligible heat capacity .
HI(aq) + KOH(aq) --> KI(aq) + H2O(l)
enthalpy of H ?
You have 65.0 g soln from HI and 84.0 g H2O with the KOH so total mass soln is 149 g.
q = mass H2O x specific heat water x (delta T).
USUALLY delta H for a reaction is done in kJ/mol. This problem has no easy way to get to moles but q/mol = J/mol and you can convert that to kJ/mol if needed. If you assume the density of the HI solution is 1.00 g/mL, that can be used to solve for mols HI; e.g., 65 mL of 0.600 M HI would be moles = M x L = 0.600 x 0.065 = ? and you can use that for mols. I doubt that the density is 1. g/mL.
To calculate the enthalpy of the reaction, we need to use the equation:
q = m·C·ΔT
where:
q = heat gained or lost by the solution (in this case, gained)
m = mass of the solution (in grams)
C = specific heat of the solution (in J/g(°C))
ΔT = change in temperature (final temperature - initial temperature)
To solve this problem, we need to break it down into steps:
Step 1: Calculate the initial temperature of the mixture
Step 2: Calculate the heat gained or lost by the solution
Step 3: Rearrange the equation to find the enthalpy change (ΔH)
Step 1: Calculate the initial temperature of the mixture
Since both the HI and KOH solutions are at 18.46 °C initially, the initial temperature of the mixture is also 18.46 °C.
Step 2: Calculate the heat gained or lost by the solution
We can calculate the heat gained or lost by the solution using the equation:
q = m·C·ΔT
For the HI solution:
q1 = (65.0 g) · (4.184 J/g°C) · (21.96 - 18.46) °C
For the KOH solution:
q2 = (84.0 g) · (4.184 J/g°C) · (21.96 - 18.46) °C
Step 3: Rearrange the equation to find the enthalpy change (ΔH)
Since the heat lost by KOH is equal to the heat gained by HI in an exothermic reaction, we can calculate the enthalpy change (ΔH) as follows:
q1 = -q2 = ΔH
ΔH = -[(65.0 g) · (4.184 J/g°C) · (21.96 - 18.46) °C]
Thus, the enthalpy change (ΔH) for the reaction is equal to -[calculated value] J.
To calculate the enthalpy change (ΔH) for the reaction, you'll need to use the equation:
q = m · C · ΔT
where:
q is the heat absorbed or released by the reaction,
m is the mass of the solution,
C is the specific heat capacity of the solution, and
ΔT is the change in temperature.
First, calculate the heat absorbed or released by the reaction (q) using the equation:
q = m1 · C1 · ΔT1 + m2 · C2 · ΔT2
where:
m1 is the mass of the HI solution,
m2 is the mass of the KOH solution,
C1 is the specific heat capacity of the HI solution, and
C2 is the specific heat capacity of the KOH solution.
Given:
m1 = 65.0 g
m2 = 84.0 g
C1 = C2 = 4.184 J/g(°C)
ΔT1 = ΔT2 = (final temperature - initial temperature)
In this case, the initial temperature for both solutions is 18.46 °C, and the final temperature is 21.96 °C.
Now you can calculate ΔT1 and ΔT2:
ΔT1 = 21.96 °C - 18.46 °C = 3.50 °C
ΔT2 = 21.96 °C - 18.46 °C = 3.50 °C
Substituting the values into the equation for q:
q = (65.0 g) · (4.184 J/g(°C)) · (3.50 °C) + (84.0 g) · (4.184 J/g(°C)) · (3.50 °C)
Calculate the value of q, the heat absorbed or released by the reaction.
Next, we need to convert q to moles of HI. First, calculate the number of moles of HI using the given mass and molar mass of HI (127.91 g/mol). The number of moles can be calculated using the equation:
moles of HI = mass of HI / molar mass of HI
Once you have the moles of HI, you can calculate the enthalpy change (ΔH) for the reaction using the equation:
ΔH = q / moles of HI
Substitute the calculated values into the equation to find the enthalpy change for the reaction.
Awesome
I am defiantly starting to understand this stuff better.
.065L x (.600 mol/L) = 0.0390 mol
q=MC(delta T)
= 149g x 4.184 J/g(C) x 3.5 celsius
= 2181.9 J
=2.2 kJ
Then I did
2.2 kJ / 0.0390 mol
= 56 kJ/mol
for my final answer I got 56 kJ/mol