Question

Practice Exercise 1 When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, the following reaction occurs: Mg(s) + 2 HCl(aq) + MgCl(aq) + H (9) If the temperature of the solution increases from 23.0 to 34.1' C as a result of this reaction, calculate AH in kJ/mol Mg. Assume that the solution has a specific heat of 4.18 J/g-° C. (a) -19.1 kJ/mol (b)-111 kJ/mol (C) -191 kJ/mol (d) -464 kJ/mol (e)-961 kJ/mol

Answer 1

Given values:

Mass of Mg = 0.243g

Volume of solution = 100mL

Solution specific heat = 4.18J / g.oC

Solution temperature; Initial (Ti) = 23.0oC, Final (Tf) = 34.1oC

Solution:

At constant pressure, the change in enthalpy can be measured as $\Delta$ H = q where $\Delta$ H is change in enthalpy and q is energy

We know that q = mc$\Delta$T --------> 1

Where q is energy; m is mass of solution = 100g (Density of water, 1g/mL is considered); c is specific heat = 4.18J / g.oC; $\Delta$ T is change in temperature (Tf - Ti) = 11.1oC

Substitute the values in equation 1 and solve,

q = 100g x (4.18J / g.oC) x 11.1oC = 4639.8J

We know that 0.243g of Mg is reacted with enough HCl to make 100 mL solution, the number moles (n) of Mg = Mass (m) of Mg / Molar mass (M) of Mg

n = 0.243g / 24.305g/mol = 0.0099979mol = 0.01mol

Now calculate the $\Delta$ H value in kJ/mol.

$\Delta$H = q / n = 4639.8J / 0.01mol = 463980J/mol

Converting in J/mol to kJ/mol,

$\Delta$H = 463980J/mol x (1 kJ / 1000J) = 463.980kJ/mol =464kJ/mol

Temperature of the surrounding is increased due to the release of heat from the reaction. Therefore the sign for $\Delta$ H is - ve and the value is - 464kJ/mol

Hence the option (d)  - 464kJ/mol is correct.