a 12.0g of sample A (70.2g/mol)was burned in a bomb calorimeter in an excess of O2(g).
A+2 O2(g)---> 2 CO2(g) +2 H2O(l)
The temperature of the calorimeter rose from 25.00 degrees celisus to 29.82 degree celisus. ifthe heat capacity of the calorimeter and its contents is 13.43kJ/degree celisus, what is the heat of change for the reaction?
q = heat of rxn = [Ccal x (Tfinal-Tintial)]
Then q/12.0 g = delta H/gram. Convert to kJ/mol by (dH/gram)*70.2 = ?
To find the heat of change for the reaction, we can use the equation:
ΔH = q / n
where ΔH is the heat of change, q is the heat released or absorbed, and n is the number of moles.
First, let's calculate the heat (q) released or absorbed by the calorimeter and its contents using the equation:
q = C × ΔT
where q is the heat, C is the heat capacity, and ΔT is the change in temperature.
Given:
C = 13.43 kJ/°C
ΔT = 29.82°C - 25.00°C = 4.82°C
Substituting the values into the equation:
q = 13.43 kJ/°C × 4.82°C
≈ 64.78 kJ
Next, we need to calculate the number of moles (n) involved in the reaction. We can use the molar mass and the mass of sample A.
Given:
Mass of sample A (m) = 12.0 g
Molar mass of sample A (M) = 70.2 g/mol
n = m / M
= 12.0 g / 70.2 g/mol
≈ 0.171 mol
Finally, we can calculate the heat of change (ΔH) using the equation:
ΔH = q / n
= 64.78 kJ / 0.171 mol
≈ 378.91 kJ/mol
Therefore, the heat of change for the reaction is approximately 378.91 kJ/mol.