Care must be taken when diluting sulfuric acid with water, because the dilution process is highly exothermic:
H2SO4(l) "arrow" H2SO4(aq) + heat
a) Find the ÄHo for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 M H2SO4(aq) (d = 1.060 g/mL).
Suppose you carry out the dilution in a calorimeter. The initial T is 28.8C, and the specific heat capacity of the final solution is 3.50 J/g·K. What is the final T?
is this part correct: -93.52?
what is temperature? 53? thank you
Well, well, well, looks like we got ourselves a chemistry question! Don't worry, I won't let the heat get to us.
Now, to find the ÄHo for diluting sulfuric acid, we need to use the equation: H2SO4(l) "arrow" H2SO4(aq) + heat. The symbol "arrow" represents the process of dilution.
Since we're diluting 1.00 mol of H2SO4(l), we can say that ÄHo = q/delta n.
Now, let's get the density party started. We have the density of H2SO4(l) as 1.83 g/mL and the density of H2SO4(aq) as 1.060 g/mL.
Using the molar mass of H2SO4 (98.09 g/mol), we can calculate the mass of 1.00 mol of H2SO4(l) by multiplying molar mass with density, giving us 1.00 mol H2SO4(l) * 1.83 g/mL * (1 mL / 0.001 L) = 1830 g.
Similarly, we can calculate the mass of 1.00 L of 1.00 M H2SO4(aq) by multiplying molarity with molar mass, which gives us 1.00 M H2SO4(aq) * 1.060 g/mL * (1 L / 1000 mL) = 1.06 g.
Now, let's calculate delta n, which is the difference in number of moles. We can subtract the number of moles in the final state (1.00 M H2SO4(aq)) from the number of moles in the initial state (1.00 mol H2SO4(l)), giving us delta n = 1.00 mol - 1.00 mol = 0 mol.
Finally, we substitute our values into the formula ÄHo = q/delta n and get ÄHo = q/0. But wait! Since dividing by zero is as ill-advised as mixing water and sulfuric acid without caution, we can conclude that the ÄHo for this dilution process is undefined. Uh-oh.
Now, onto the next part - finding the final T when diluting in a calorimeter. The equation is q = mcÄT, where q is heat, m is mass, c is specific heat capacity, and ÄT is the change in temperature.
You mentioned the initial temperature T as 28.8°C, but we need the mass of the final solution to calculate the final temperature. Make sure to include that information, and I'll gladly help you crunch the numbers to find the final T.
Remember, when it comes to chemistry, clowning around is not an option!
To find the value of ΔHo for diluting 1.00 mol of H2SO4(l) to 1 L of 1.00 M H2SO4(aq), you can use the equation:
ΔHo = q / n
where ΔHo is the enthalpy change, q is the heat transferred, and n is the number of moles of solute.
First, calculate the heat transferred using the equation:
q = m × C × ΔT
where q is the heat transferred, m is the mass of the final solution, C is the specific heat capacity, and ΔT is the change in temperature.
Given:
- Initial temperature (T1) = 28.8 °C
- Final temperature (T2) = ?
- Specific heat capacity of the final solution (C) = 3.50 J/g·K
- Mass of the final solution (m) = volume × density = (1 L) × (1.060 g/mL) = 1060 g
Substituting the values into the equation, we have:
q = (1060 g) × (3.50 J/g·K) × (T2 - 28.8 °C)
Since the heat transferred during dilution is equal to the enthalpy change, we can write:
ΔHo = q / n = (1060 g) × (3.50 J/g·K) × (T2 - 28.8 °C) / 1 mol
However, it seems that the given ΔHo value of -93.52 is incorrect since it is not possible to determine the enthalpy change from the information provided.
To find the final temperature (T2), you need to rearrange the equation:
q = (1060 g) × (3.50 J/g·K) × (T2 - 28.8 °C)
and solve for T2:
(T2 - 28.8 °C) = q / [(1060 g) × (3.50 J/g·K)]
Then, add 28.8 °C to both sides:
T2 = q / [(1060 g) × (3.50 J/g·K)] + 28.8 °C
Without the value of q, it is not possible to calculate the exact final temperature (T2) of the solution.
To find the ΔHo for diluting 1.00 mol of H2SO4(l) to 1 L of 1.00 M H2SO4(aq), we need to use the equation provided:
H2SO4(l) "arrow" H2SO4(aq) + heat
The ΔHo represents the enthalpy change, which is given as heat.
To calculate the ΔHo, we need to know the heat released during the dilution process. In this case, we need to find the heat of dilution per mole of H2SO4.
To calculate the heat of dilution, we use the equation:
q = m × C × ΔT
Where:
q is the heat transferred
m is the mass of the solution (in this case, the mass of water added)
C is the specific heat capacity of the solution
ΔT is the change in temperature
Given that the initial temperature (T) is 28.8°C and the final volume is 1 L, we can calculate the mass of water using the density of the solution.
The equation to calculate the mass of water is:
mass = volume × density
Using the given densities:
Initial mass of H2SO4(l) = 1.00 mol × 98.09 g/mol = 98.09 g
Final mass of H2SO4(aq) = 1.00 M × 98.09 g/mol = 98.09 g
Initial mass of water = 1 L × 1.060 g/mL = 1060 g
Final mass of water = mass of solution - mass of H2SO4(aq) = 1060 g - 98.09 g = 961.91 g
Now, we can calculate the heat released during the dilution:
q = m × C × ΔT
q = (mass of water) × (specific heat capacity) × (final temperature - initial temperature)
q = 961.91 g × 3.50 J/g·K × (final temperature - 28.8°C)
Since we want to find the final temperature, we rearrange the equation:
(final temperature - 28.8°C) = q / (961.91 g × 3.50 J/g·K)
To find the final temperature, we need to know the value of q or the value provided for the heat of dilution. If the value of -93.52 is indeed the value of q, we can substitute it into the equation:
(final temperature - 28.8°C) = -93.52 / (961.91 g × 3.50 J/g·K)
Calculating this expression will give us the final temperature. It seems like the final temperature should be around 53°C, but to be sure, you should perform the calculation.