When 15.0 g of sodium nitrate (NaNO3) was dissolved in water in a bomb calorimeter, the temperature reduced from 25.00 °C to 21.56 °C. If the heat capacity of the solution and the calorimeter is 1071kJ/°C, what is the enthalpy change for the process per mol of sodium nitrate that dissolves in water? (Atomic masses of N-14.007 g/mol; Na-22.990 g/mol;O-15.999 g/mol)
-13.03x10 to the power of 4 kJ/mol
+13684.24 x10 to the power of 4 kJ/mol
-52.99x10 to the power of 4 kJ/mol
-2.09x10 to the power of 4 kJ/mol
The correct answer is -2.09x10 to the power of 4 kJ/mol.
First, calculate the amount of heat released during the reaction:
q = mcΔT
q = (1071 kJ/°C)(3.44 °C)(0.0150 kg) (convert grams to kg)
q = - 52.99 kJ
Note that the "minus" sign in the answer indicates that heat is released during the reaction.
Next, calculate the moles of NaNO3:
moles NaNO3 = mass/molar mass = 15.0 g/(22.990 g/mol + 14.007 g/mol + 3(15.999 g/mol)) = 0.0881 mol
Finally, calculate the enthalpy change per mole of NaNO3:
ΔH = q/moles of NaNO3 = -52.99 kJ/0.0881 mol = -2.09x10 to the power of 4 kJ/mol
To find the enthalpy change per mole (ΔH) for the process of dissolving sodium nitrate in water, we can use the equation:
ΔH = q / n
where q is the heat absorbed or released by the system (in this case, the calorimeter and solution) and n is the number of moles of sodium nitrate dissolved.
First, we need to calculate the heat absorbed or released by the system (q). We can use the equation:
q = C * ΔT
where C is the heat capacity of the solution and the calorimeter and ΔT is the change in temperature. Given that C = 1071 kJ/°C and ΔT = (21.56 °C - 25.00 °C) = -3.44 °C, we can calculate q:
q = 1071 kJ/°C * (-3.44 °C) = -3680.24 kJ
Next, we need to calculate the number of moles of sodium nitrate (NaNO3). The molar mass of NaNO3 is:
Na: 22.990 g/mol
N: 14.007 g/mol
O: 15.999 g/mol
NaNO3: 22.990 g/mol + 14.007 g/mol + 3 * 15.999 g/mol = 85.01 g/mol
Given that the mass of sodium nitrate dissolved is 15.0 g, we can calculate the number of moles:
n = 15.0 g / 85.01 g/mol = 0.176 mol
Finally, we can substitute the values into the equation for ΔH:
ΔH = -3680.24 kJ / 0.176 mol = -20,909.52 kJ/mol
Rounded to the appropriate number of significant figures, the enthalpy change per mol of sodium nitrate that dissolves in water is approximately -2.09 × 10^4 kJ/mol. Therefore, the correct answer is -2.09 × 10^4 kJ/mol.