In a constant-pressure calorimeter, 60.0 mL of 0.330 M Ba(OH)2 was added to 60.0 mL of 0.660 M HCl. The reaction caused the temperature of the solution to rise from 21.71 °C to 26.21 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.I need help asap!!! answer in kJ/mol H2O
I've just gone through this with Akle below Here is a link. The numbers are different but the method is ok. Post your work here if you have trouble and I'll try to find the error.
http://www.jiskha.com/display.cgi?id=1328928607
Oh dear, looks like you're in hot water with this question! But don't worry, I'll do my best to help you out.
To find the enthalpy change (ΔH) for this reaction per mole of H2O produced, we'll need to use the equation:
ΔH = q / n
Where:
- ΔH is the enthalpy change
- q is the heat absorbed or released by the reaction
- n is the number of moles of H2O produced
First, let's calculate the heat absorbed or released by the reaction using the formula:
q = m * c * ΔT
Where:
- q is the heat absorbed or released by the reaction
- m is the mass of the solution (which we'll assume is equal to the volume, since the density and specific heat are the same as water)
- c is the specific heat of water
- ΔT is the change in temperature
Since the total volume is the sum of the individual volumes, the mass of the solution is 60.0 mL + 60.0 mL = 120.0 mL.
Now, let's plug in the values:
q = 120.0 mL * 4.18 J/g°C * (26.21 °C - 21.71 °C)
I'll let you do the math to calculate q, and remember to convert the answer from J to kJ.
Once you have q, you'll also need to determine the number of moles of H2O produced. Based on the balanced chemical equation for the reaction, you should be able to determine the stoichiometric ratio between H2O and Ba(OH)2 (or HCl).
Finally, divide q by the number of moles of H2O to find ΔH.
I hope that helps you tackle this problem! Just remember to keep your cool and break it down step-by-step.
To calculate the enthalpy change (ΔH) of the reaction per mole of H2O produced, we can use the equation:
ΔH = q / n
where q is the heat gained or lost by the system, and n is the number of moles of H2O produced.
To find the heat gained or lost by the system, we can use the equation:
q = m × C × ΔT
where m is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's calculate the mass of the solution. Since the density and volume of the solution are the same as water, the mass of the solution will be the sum of the masses of the individual solutions:
mass = volume × density = (60.0 mL + 60.0 mL) × 1 g/mL = 120.0 g
Now, let's calculate the change in temperature:
ΔT = final temperature - initial temperature = 26.21 °C - 21.71 °C = 4.50 °C
The specific heat capacity of water (C) is 4.18 J/g°C. To convert this to kJ/mol°C, we need to use the molar mass of water, which is approximately 18.02 g/mol. Therefore, the specific heat capacity in kJ/mol°C is:
C = 4.18 J/g°C × (1 kJ/1000 J) × (1 mol/18.02 g) ≈ 0.231 kJ/mol°C
Now, let's calculate the heat gained or lost by the system:
q = (120.0 g) × (0.231 kJ/mol°C) × (4.50 °C) = 12.377 kJ
Finally, we can calculate the enthalpy change per mole of H2O produced:
n(H2O) = moles of limiting reactant
In this case, the limiting reactant is Ba(OH)2, since its concentration is lower than that of HCl. The balanced chemical equation for the reaction is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
From the balanced equation, we can see that 2 moles of H2O are produced per mole of Ba(OH)2 reacted.
n(H2O) = (0.060 L) × (0.330 mol/L) × (2 mol H2O / 1 mol Ba(OH)2) = 0.0396 mol
Therefore, ΔH per mole of H2O produced is:
ΔH = q / n = 12.377 kJ / 0.0396 mol ≈ 312.22 kJ/mol H2O
So, ΔH for this reaction per mole of H2O produced is approximately 312.22 kJ/mol H2O.
To calculate the enthalpy change (ΔH) per mole of H2O produced in this reaction, several steps need to be followed:
1. Calculate the heat absorbed or released by the solution using the equation:
q = m * c * ΔT
Where:
- q is the heat absorbed or released by the solution
- m is the mass of the solution
- c is the specific heat capacity of the solution (assumed to be the same as water)
- ΔT is the change in temperature
Since the density of the solution is the same as water, the mass (m) can be calculated as:
m = V * d
Where:
- V is the volume of the solution (sum of individual volumes)
- d is the density of the solution (assumed to be the same as water)
2. Calculate the number of moles of Ba(OH)2 and HCl that reacted:
Moles of Ba(OH)2 = volume (in L) * concentration
Moles of HCl = volume (in L) * concentration
3. Determine the limiting reactant. The reactant that produces fewer moles of product is the limiting reactant. The stoichiometry of the balanced chemical equation can be used to determine the ratio of moles of H2O produced to moles of reactant consumed.
4. Calculate the enthalpy change per mole of H2O produced:
ΔH = q / moles of H2O
Now let's go through the calculations:
Step 1: Calculating the heat absorbed or released by the solution (q)
- ΔT = (26.21 °C - 21.71 °C) = 4.5 °C
- V = 60.0 mL + 60.0 mL = 120.0 mL = 0.120 L
- d (density of water) = 1.00 g/mL = 1.00 g/cm³
- m = V * d = 0.120 L * 1.00 g/cm³ = 0.120 g
- c (specific heat capacity of water) = 4.18 J/g°C (or 4.18 kJ/kg°C)
Now, calculate q using the formula:
q = m * c * ΔT = 0.120 g * 4.18 kJ/kg°C * 4.5°C = 0.226 kJ
Step 2: Calculating moles of Ba(OH)2 and HCl
- Moles of Ba(OH)2 = 0.060 L * 0.330 mol/L = 0.0198 mol
- Moles of HCl = 0.060 L * 0.660 mol/L = 0.0396 mol
Step 3: Determining the limiting reactant
The balanced chemical equation for the reaction between Ba(OH)2 and HCl is:
Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + 2H2O(l)
According to the stoichiometry, one mole of Ba(OH)2 reacts with 2 moles of HCl to produce 2 moles of H2O. Since the stoichiometric ratio is 1:2, the limiting reactant is Ba(OH)2 since it produces fewer moles of product.
Step 4: Calculating the enthalpy change per mole of H2O produced
- Since Ba(OH)2 is the limiting reactant, the number of moles of H2O produced is equal to the number of moles of Ba(OH)2.
- ΔH = q / moles of H2O = 0.226 kJ / 0.0198 mol = 11.4 kJ/mol H2O
Therefore, the enthalpy change (ΔH) per mole of H2O produced in this reaction is 11.4 kJ/mol H2O.