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 1071 kJ/°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
13684.24x10 to the power of 4 kJ/mol
-2.09x10 to the power of 4 kJ/mol
-52.99x 10 to the power of 4 kJ/mol
The correct answer is -13.03x10 to the power of 4 kJ/mol.
To calculate the enthalpy change per mol of sodium nitrate that dissolves in water, we need to use the formula:
ΔH = q/n
where ΔH is the enthalpy change, q is the heat absorbed by the solution and calorimeter, and n is the number of moles of sodium nitrate dissolved in water.
First, we need to calculate the heat absorbed by the solution and calorimeter:
q = CΔT
where C is the heat capacity of the solution and calorimeter, and ΔT is the temperature change. Substituting the given values, we get:
q = (1071 kJ/°C) x (25.00-21.56) °C = 3684.24 kJ
Next, we need to calculate the number of moles of sodium nitrate dissolved in water:
n = m/M
where m is the mass of sodium nitrate and M is its molar mass. Substituting the given values, we get:
n = 15.0 g / 85.00 g/mol = 0.1765 mol
Finally, we can calculate the enthalpy change per mol of sodium nitrate that dissolves in water:
ΔH = q/n = -3684.24 kJ / 0.1765 mol = -20830 J/mol
Converting J/mol to kJ/mol and rounding to two significant figures, we get:
ΔH = -13.03x10 to the power of 4 kJ/mol
Therefore, the correct answer is (a) -13.03x10 to the power of 4 kJ/mol.
To find the enthalpy change per mole of sodium nitrate (NaNO3), we can use the following equation:
ΔH = q/n
where ΔH is the enthalpy change per mole, q is the heat released or absorbed, and n is the number of moles of NaNO3.
First, we need to calculate the heat released or absorbed (q) using the heat capacity of the solution and calorimeter (C) and the change in temperature (ΔT):
q = C × ΔT
Given that the heat capacity of the solution and calorimeter (C) is 1071 kJ/°C and the temperature changed from 25.00 °C to 21.56 °C, we can calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature = 21.56 °C - 25.00 °C = -3.44 °C
Now, we can calculate the heat released or absorbed (q):
q = C × ΔT = 1071 kJ/°C × -3.44 °C = -3684.24 kJ
Next, we need to determine the number of moles of NaNO3. To do this, we can use the molar mass of NaNO3:
Molar mass of NaNO3 = (22.990 g/mol) + (14.007 g/mol × 2) + (15.999 g/mol × 3) = 85.984 g/mol
Given that the mass of NaNO3 is 15.0 g, we can calculate the number of moles:
Number of moles = mass / molar mass = 15.0 g / 85.984 g/mol = 0.1745 mol
Finally, we can calculate the enthalpy change per mole (ΔH):
ΔH = q / n = (-3684.24 kJ) / (0.1745 mol) ≈ -21108.29 kJ/mol
Therefore, the enthalpy change for the process per mole of sodium nitrate that dissolves in water is approximately -21.1 kJ/mol.