The heat of neutralization of HCl (aq) by NaOH is -55.9Kj/mol H2) produced. If 50ml of 1.6M NaOH at 25.15 celsius is added to 25ml of 1.79M HCl at 26.34 celsius in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occured?
I'm confused by the term "immediately after the neutralization reaction has occured?" Does that mean we aren't giving the solution time to reach equilibrium temperature wise. The second thing that bothers me about this problem is that you do NOT have any H2 gas released.
NaOH + HCl ==> NaCl + H2O
To determine the solution temperature immediately after the neutralization reaction, we can apply the principle of heat exchange. The heat released or absorbed during the reaction can be calculated using the following equation:
q = m * c * ΔT
Where:
q = heat released or absorbed (in joules)
m = mass of the solution (in grams)
c = specific heat capacity of the solution (in J/g·°C)
ΔT = change in temperature (in °C)
First, let's find the mass of the solution:
For the NaOH solution:
volume = 50 mL = 50 cm³
density of water = 1 g/cm³
mass = volume * density = 50 g
For the HCl solution:
volume = 25 mL = 25 cm³
density of water = 1 g/cm³
mass = volume * density = 25 g
Now, let's find the change in temperature (ΔT):
ΔT = final temperature - initial temperature
Since the initial and final temperatures are given, we can calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 25.15°C - 26.34°C
ΔT = -1.19°C
Next, we need to calculate the total heat released during the neutralization reaction:
q = mNaOH * cNaOH * ΔTNaOH + mHCl * cHCl * ΔTHCl
To calculate the heat released by NaOH (mNaOH * cNaOH * ΔTNaOH), we need the specific heat capacity (cNaOH) of the NaOH solution. The specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C.
The specific heat capacity of NaOH is not usually provided, so we can assume it is similar to the specific heat capacity of water, which is approximately 4.18 J/g·°C.
Using the given values:
mNaOH = 50 g
cNaOH = 4.18 J/g·°C
ΔTNaOH = -1.19°C
qNaOH = mNaOH * cNaOH * ΔTNaOH
qNaOH = 50 g * 4.18 J/g·°C * -1.19°C
Now let's calculate the heat released by HCl:
To calculate the heat released by HCl (mHCl * cHCl * ΔTHCl), we need the specific heat capacity (cHCl) of the HCl solution. Again, the specific heat capacity of HCl is usually not provided, so we can assume it is similar to the specific heat capacity of water, which is approximately 4.18 J/g·°C.
Using the given values:
mHCl = 25 g
cHCl = 4.18 J/g·°C
ΔTHCl = -1.19°C
qHCl = mHCl * cHCl * ΔTHCl
qHCl = 25 g * 4.18 J/g·°C * -1.19°C
Finally, let's calculate the total heat released (qTotal) during the neutralization reaction:
qTotal = qNaOH + qHCl
Now that we have calculated the total heat released during the neutralization reaction, we can calculate the change in temperature (ΔT) using the equation:
qTotal = mTotal * cTotal * ΔT
Rearranging the equation, we can solve for ΔT:
ΔT = qTotal / (mTotal * cTotal)
Now we have all the necessary information to calculate the solution temperature immediately after the neutralization reaction has occurred.