when 21.70 ml of 0.5000 m h2so4 is added to 21.70 mL of 1.000 M KOH in a coffee cup calorimeter at 23.50 degrees c, the temperature rises to 30.17 degree c. calculate delta H of this reaction (Assume that the total volume is the sum of the individual volumes and that density and specific heat capacity of the solution are the same as pure water d of water =1.00g/mL: c of water =4.184J/g*c) I have no idea where to even start
Well, it seems like you're in a bit of a stew with this problem. But fear not, I'll help you brew up a solution.
To find the enthalpy change (ΔH) of this reaction, we can use the equation:
q = m × c × ΔT
Where:
q is the heat gained or lost by the reaction
m is the mass of the solution (which we can find using the density of water)
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's find the mass of the solution. Since the density of water is given as 1.00 g/mL, we know that the mass of 21.70 mL of the solution is 21.70 g.
Now, let's calculate the heat gained or lost by the reaction. Since the reaction is exothermic (the temperature increased), the heat value will be negative to indicate that energy was released.
q = -m × c × ΔT
q = -21.70 g × 4.184 J/g°C × (30.17°C - 23.50°C)
Now, let's plug in the values and calculate:
q = -21.70 g × 4.184 J/g°C × 6.67°C
Solving this gives us:
q ≈ -623.6 J
Finally, we know that ΔH is equal to q (since the reaction is done at constant pressure). Therefore:
ΔH = -623.6 J
So, the enthalpy change (ΔH) of this reaction is approximately -623.6 J. It's negative because energy was released, which is typical for an exothermic reaction.
To calculate the enthalpy change (∆H) of the reaction, we can use the equation:
∆H = q / n
Where:
q = heat transferred in the reaction
n = number of moles of the limiting reactant
To determine the heat transferred, we can use the equation:
q = m × c × ∆T
Where:
m = mass of the solution
c = specific heat capacity of water
∆T = change in temperature
Step 1: Determine the mass of the solution.
We know that the density of water is 1.00 g/mL.
So, the mass of the solution is equal to its volume.
mass of solution = volume of solution = 21.70 mL + 21.70 mL
Step 2: Calculate the heat transferred (q).
We will use the equation q = m × c × ∆T.
Step 3: Calculate the number of moles of the limiting reactant.
We can use the molarity (M) and volume (V) relationship to calculate the number of moles (n) of the limiting reactant.
n(moles) = M × V(L) = M × V(mL) / 1000
Step 4: Calculate the enthalpy change (∆H).
Using the equation ∆H = q / n, substitute the values obtained from the previous steps.
Now let's calculate each step:
Step 1:
mass of solution = volume of solution = 21.70 mL + 21.70 mL = 43.40 g
Step 2:
q = m × c × ∆T
q = 43.40 g × 4.184 J/g°C × (30.17°C - 23.50°C)
Step 3:
n(moles) = M × V(mL) / 1000
n(H2SO4) = 0.5000 mol/L × 21.70 mL / 1000 = 0.01085 moles
Step 4:
∆H = q / n
∆H = q / n(H2SO4)
By substituting the values obtained for q and n(H2SO4), you can calculate the enthalpy change (∆H) of the reaction.
To calculate the enthalpy change, ΔH, of a reaction, you can use the equation:
ΔH = q / n
Where:
q is the heat absorbed or released in the reaction
n is the number of moles of the limiting reactant
To calculate q, the heat absorbed or released, you can use the equation:
q = m × c × ΔT
Where:
m is the mass of the solution
c is the specific heat capacity of water (assumed to be the same as the solution)
ΔT is the change in temperature
First, calculate the mass of the solution:
The total volume of the solution is the sum of the volumes of the H2SO4 and KOH solutions. Since the density of water is 1.00 g/mL, the mass can be calculated using:
mass = volume × density
mass = (21.70 mL + 21.70 mL) × (1.00 g/mL)
Next, calculate the change in temperature, ΔT:
ΔT = final temperature - initial temperature
ΔT = 30.17 °C - 23.50 °C
Now, calculate q:
q = mass × c × ΔT
q = (mass) × (4.184 J/g*°C) × (ΔT)
Finally, calculate ΔH:
Since H2SO4 and KOH react in a 1:1 ratio, the number of moles of each is the same. Therefore, the number of moles of the limiting reactant is equal to the number of moles of either H2SO4 or KOH. You can calculate the moles using the formula:
moles = concentration × volume
The concentration of H2SO4 is 0.5000 M, so the number of moles is:
moles = (0.5000 M) × (21.70 mL)
The ΔH can now be calculated as:
ΔH = q / n
Substitute the calculated values of q and n into the equation to find the enthalpy change, ΔH.
2KOH + H2SO4 ==> 2H2O + K2SO4
mols KOH = M x L = 0.0217
mols H2SO4 = 0.01085
mols KOH needed = 2*0.01085 = 0.02170g so the solutions exactly neutralize each other.
mols H2O formed = 0.02170
volume = 21.70 + 21.70 = 43.40 mL
g H2O = 43.40 grams
qrxn = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 43.40 g x 4.184 J/g*C x(30.17-23.50)
q = about 1210 J but you need a better number than this estimate.
About 1210 J for 0.0217 mol is
about 1210/0.0217 = about 55 kJ/mol