Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic reaction: NH_4NO_3(s) rightarrow NH_4 ^ +(aq)+ NO_3 ^ -(aq)
In order to measure the enthalpy change for this reaction, 1.25g of NH_4NO_3 is dissolved in enough water to make 25.0mL of solution. The initial temperature is 25.8 ^\circ {\rm C} and the final temperature (after the solid dissolves) is 21.9 ^\circ {\rm C}. Calculate the change in enthalpy for the reaction. (Use 1.0g/mL as the density of the solution and 4.18J/g * {\rm{^\circ C}}} as the specific heat capacity.)
Savannah, you could save your self a huge amount of time by leaving all of those funny symbols out.
q = delta H = (mass H2O x specific heat H2O x (Tfinal-Tinitial)
q/g NH4NO3 = q/1.25 = dH in joules.
delta H in J/mol = (q/1.25)*molar mass NH4NO3. Then change to kJ/mol; that's the usual way of reporting.
i tried using that formula and keep getting it wrong but i don't know why.
Show your work so we can see what you did.
The mass of the water is 25.0 mL x 1.0 g/mL = 25.0 g and not 18g. That will make a difference.
The molar mass of NH4NO3 is 80.04 and not 80.6; that will make a difference.
Finally, if you reported the answer to as many places as you show then the data base probably is telling you that you are reporting too many significant figures. You are allowed only three. If your answer were correct you would be allowed 18.9 kJ/mol.
I just thought of another point. In your post there is nothing about which unit to use for reporting; look over the problem and make sure you are using the correct unit(s).
Hope this works for you.
To calculate the change in enthalpy (ΔH) for the reaction, you can use the equation:
ΔH = q / n,
where q is the amount of heat transferred in the reaction and n is the number of moles of NH4NO3.
First, let's calculate q using the equation:
q = mcΔT,
where m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
Since the density of the solution is given as 1.0 g/mL, we can calculate the mass of the solution using the formula:
mass = volume × density,
mass = 25.0 mL × 1.0 g/mL,
mass = 25.0 g.
Now we can calculate q:
q = (25.0 g) × (4.18 J/g * °C) × (21.9 °C - 25.8 °C),
q = -323.85 J.
Since 1.25 g of NH4NO3 is dissolved in the solution, we can calculate the number of moles of NH4NO3:
n = mass / molar mass,
where the molar mass of NH4NO3 is:
NH4NO3 = (1 × 14.01 g/mol) + (4 × 1.01 g/mol) + 3 × (16.00 g/mol),
NH4NO3 = 80.04 g/mol.
n = 1.25 g / 80.04 g/mol,
n = 0.01562 mol.
Now we can calculate the change in enthalpy (ΔH):
ΔH = q / n,
ΔH = -323.85 J / 0.01562 mol,
ΔH = -20760 J/mol.
Therefore, the change in enthalpy (ΔH) for the reaction is -20760 J/mol.
18.02g*4.18J/g*C*-3.9=-293.76204
-293.76204/1.25=-235.009632
-235.009632*80.6=-18941.77634J/mol
-18941.77634J/mol=-18.94177634kJ/mol