1) A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/C. The temperature increases by 1.126 degrees C. Calculate the heat given off by the burning Mg, in kJ/g and kJ/mol.
2) Determine the amount of heat in kJ given off when 1.26 x 10^4 g of NO2 are produced according to the equation:
2NO (g) + O2 (g) --> 2NO2 (g)
and delta (H) = -114.6 kJ/mol
1.
q = Ccal*(Tfinal-Tinitial) for the rxn.
q/0.1375 = q/gram
q/gram x atomic mass mg = q/mol. Change those to kJ/g and kJ/mol
A 6.22-kg piece of copper metal is heated from 20.5°C to 324.3°C. Calculate the heat absorbed (in kJ) by the metal. sCu= 0.385 J/ g oC
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/°C. The temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol. s Mg = 1.020 J/ g oC
1) Oh boy, time to burn some magnesium! But before we light things up, let's calculate the heat given off.
To convert the temperature change to joules, we multiply it by the heat capacity of the bomb calorimeter:
ΔQ = (1.126 °C) * (3024 J/C)
ΔQ = 3404.224 J
Now, let's convert the mass of magnesium to moles. The molar mass of magnesium (Mg) is 24.31 g/mol:
moles of Mg = (0.1375 g) / (24.31 g/mol)
moles of Mg = 0.00566 mol
Finally, let's divide the heat change by the mass of magnesium to get the heat given off per gram:
Heat given off in kJ/g = (3404.224 J) / (0.1375 g) => *(Hint: remember to convert J to kJ!)*
2) Now we're talking about NO2! According to the equation, we need 2 moles of NO to react with 1 mole of O2 to produce 2 moles of NO2. So, to find the moles of NO2, we'll have to figure out the moles of NO first.
The mass of NO2 is 1.26 x 10^4 g. To find the moles, we need to divide the mass by the molar mass of NO2, which is 46.01 g/mol:
moles of NO2 = (1.26 x 10^4 g) / (46.01 g/mol)
Once we have the moles of NO2, we can use the stoichiometry of the balanced equation to find the moles of NO assuming all reactants are fully consumed (which is what we are assuming here). We'll need twice the amount of NO moles compared to NO2 moles:
moles of NO = 2 * moles of NO2
Finally, since the enthalpy change (delta H) is given per mole of NO2 produced, we can multiply the moles of NO2 by the enthalpy change to find the heat given off:
Heat given off in kJ = moles of NO2 * (-114.6 kJ/mol)
To solve these problems, we need to use the concept of heat capacity and the change in temperature.
For the first problem:
1) To calculate the heat given off by the burning Mg in kJ/g, we'll first convert the mass of magnesium from grams to kilograms because the heat capacity is given in J/C, not J/°C per gram. So, the mass of magnesium is 0.1375 g, which is equal to 0.0001375 kg.
2) Next, we'll multiply the mass of magnesium by the change in temperature to get ΔT. In this case, ΔT is 1.126 °C.
ΔT = 1.126 °C = 1.126 K
3) Now we'll use the formula Q = mcΔT, where Q represents the heat, m represents the mass, c represents the heat capacity, and ΔT represents the change in temperature.
Q = (0.0001375 kg) * (3024 J/C) * (1.126 K)
= 0.04724 J
4) Finally, to convert the heat from J to kJ, we divide the answer by 1000.
Q = 0.04724 J = 0.00004724 kJ
To obtain the result in kJ/g, we divide the heat by the mass of magnesium:
Q in kJ/g = (0.04724 J) / (0.1375 g) = 0.343 kJ/g
To calculate the heat given off by the burning Mg in kJ/mol, we need to know the molar mass of magnesium. The molar mass of Mg is 24.31 g/mol.
5) Now, we can calculate the number of moles of magnesium in the given sample:
Moles of Mg = (0.1375 g) / (24.31 g/mol) = 0.00566 mol
6) To find the heat in kJ/mol, we'll divide the heat obtained in step 3 by the number of moles of magnesium:
Q in kJ/mol = (0.04724 J) / (0.00566 mol) = 8.34 kJ/mol
Therefore, the heat given off by the burning magnesium is 0.343 kJ/g and 8.34 kJ/mol.
For the second problem:
1) The balanced equation gives the stoichiometric ratio between NO and NO2. From the equation, we can see that 2 moles of NO react to form 2 moles of NO2.
2) To find the moles of NO2 produced, we divide the given mass of NO2 (1.26 x 10^4 g) by the molar mass of NO2 (46.01 g/mol):
Moles of NO2 = (1.26 x 10^4 g) / (46.01 g/mol) = 273.8 mol
3) Now, we use the value of the ΔH from the balanced equation (-114.6 kJ/mol) to calculate the heat given off:
Q = (273.8 mol) * (-114.6 kJ/mol) = -31.4 kJ
Therefore, the heat given off when 1.26 x 10^4 g of NO2 are produced is -31.4 kJ. The negative sign indicates that the reaction is exothermic and gives off heat.
2.
If 114.6 kJ are produced per 1 mol NO then 2*114.6 or 229.2 kJ will be produced for 2 mols(2*30 = 60 g).
You may want to check this out in the problem because it always leaves one in doubt exactly what is meant. I'll assume the above is correct.
Then 229.2 kJ x (1.16E4g/60) = heat produced.