When 22.3 mL of 1.00 M AgNO3 is mixed with 21.7 mL of 1.50 M NaCl, the temperature of the mixture rises for 21.6 degrees C to 29.7 degrees C. The mass of the aqueous mixture is 45.9 grams and its specific heat capacity is 3.94 J/grams degrees C. I calculated (45.9g)x (3.94J/g Celsius) x (8.1 Celsius), which came out to 1464.8526 Joules. How do I use this into solving for Delta H?
Delta H is that heat divided by moles of AgCl formed....you have to calculate that.
doesnt the concentration come into play some how?
Yes and that's what Mr. Pursley told you. You must calculate the mols AgCl.
To solve for ΔH (enthalpy change) in this scenario, you can use the equation:
ΔH = mcΔT
Where:
ΔH is the enthalpy change (in Joules)
m is the mass of the mixture (in grams)
c is the specific heat capacity of the mixture (in J/grams degrees C)
ΔT is the change in temperature (in degrees C)
In your case, you have already calculated the value for mcΔT as 1464.8526 Joules. To find ΔH, you need to rearrange the equation:
ΔH = mcΔT
Plug in the known values:
ΔH = (45.9 g) x (3.94 J/g°C) x (8.1°C)
Now, you can directly calculate ΔH:
ΔH = 1504.081 Joules
Therefore, the enthalpy change (ΔH) for the given reaction is approximately 1504.081 Joules.