I have done everything to solve this problem and I can't get the right answer (153.8J/degrees C).
Ammonium nitrate, a component of fertilizer, decomposes: NH4NO3(s) > N2O(g) + 2H2) (g)
the delta H=-36.0kJ/mol.
7.65g of ammonium nitrate was allowed to decompose in a bomb calorimeter containing 415g of water. The heat produced resulted in a temperature increase of 1.82 degrees C. What is the heat capacity of the calorimeter, in J/K? The specific heat capacity of water=4.184 J/K/g
Thanks!
q(NH4NO3) + q(H2O) + q(cal) = 0
[(7.65/80.04)*36,000]+[(415*4.184*1.92)]+(C*1.82)=0
Solve for C.
I still don't get 153.8 and I am unsure as to where the 1.92 came from?
The 1.92 is my typo. It should be 1.82 from the problem. It doesn't give 153.8 but it's close. I think that's the way to do the problem.
To solve this problem, you need to follow a series of steps:
Step 1: Convert the given mass (7.65g) of ammonium nitrate to moles using its molar mass.
The molar mass of ammonium nitrate (NH4NO3) can be calculated by adding up the atomic masses of its constituent elements:
Molar mass of NH4NO3 = (1 * 14.01 g/mol) + (4 * 1.01 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol) = 80.04 g/mol
Now, divide the given mass by the molar mass to convert grams to moles:
Moles of NH4NO3 = 7.65g / 80.04 g/mol = 0.0956 mol
Step 2: Calculate the heat released by the decomposition of ammonium nitrate.
Given that the delta H is -36.0 kJ/mol and we have 0.0956 mol of NH4NO3, we can multiply the two values to find the heat released:
Heat released = (delta H) * (moles of NH4NO3)
= (-36.0 kJ/mol) * (0.0956 mol)
= -3.45 kJ
Step 3: Convert the released heat to joules.
Since the question asks for the heat capacity of the calorimeter in J/K, we need to convert the units from kilojoules to joules:
Heat released = -3.45 kJ * (1000 J/kJ)
= -3450 J
Step 4: Use the heat released to calculate the heat capacity of the calorimeter.
The heat capacity of the calorimeter (C_calorimeter) can be calculated using the following formula:
C_calorimeter = -(Heat released) / (Temperature change)
Given that the heat released is -3450 J and the temperature change is 1.82 degrees C, plug these values into the equation:
C_calorimeter = -(3450 J) / (1.82 degrees C)
= -1895.05 J/degree C
Note: The negative sign indicates that heat is being released by the system, i.e., the calorimeter.
Step 5: Finally, convert the heat capacity to J/K by dividing by the given mass of water.
Given that the specific heat capacity of water is 4.184 J/K/g and the mass of water is 415 g, we can calculate the heat capacity of the calorimeter in J/K:
Heat capacity of calorimeter = (C_calorimeter) * (mass of water)
= (-1895.05 J/degree C) * (415 g)
= -784748.75 J/K
Note: The negative sign indicates that heat is being released by the system, i.e., the calorimeter.
Hence, the heat capacity of the calorimeter is approximately -784748.75 J/K.