Question

In a calorimeter made from Iron 38.2 grams of (iodine) I2 (s) was produced by the decomposition of (hydrogen iodide) HI (g) from the data below, calculate the molar enthalpy change of I2 (s) in this reaction.

Mass of (iodine) I2(s)-----38.2 g
Mass of iron calorimeter------70.0 g
the volume of H2O(L)--------100 mL
Initial temperature--------6.0 celsius
Final temperature-------23.8 celsius

Answer 1

The number of the moles of iodine can be calculated from its mass, dividing the mass by the molar mass of iodine (253.81 g/mol):

$n(I_2)=\frac{m}{M} = \frac{38.2}{253.81}=0.1505$ mol.

The heat released in the reaction can be calculated from the change in temperature of the iron and water:

$Q = -(c_{iron}m_{iron} + c_wm_w)(T_2-T_1)$ ,

where $c$ is the specific heat capacity. Its value for water is 4.186 J/g°C and for iron is 0.450 J/g°C. Assuming the density of water equal to 1g/mL, the mass of water is 100 g.

$Q = -(0.450·100 + 4.186·70)·(23.8-6.0) = -6017$ J, or -6.0 kJ

The molar enthalpy change is the ratio of heat given off to the number of the moles:

$∆H = \frac{Q}{n}=\frac{-6017}{0.1505} = -40.0$ kJ/mol.

Answer: The molar enthalpy change of the reaction is -40.0 kJ/mol