Question

In a calorimetry experiment, 0.1277 g of Mg ribbon was added to 200.0 mL 0.500 M HCl at 24.12 °C.

The water temperature increased to 27.10 °C. Calculate ∆H per mole of HCl. (ans: –24.9 kJ/mol HCl)

Answer 1

∆H = - 15.9 kJ/mol HCl

Step-by-step explanation:

To find the amount of heat released/absorbed, (in this case, since water temperature increased, the amount released), apply the formula,

q = mc∆T, where q = amount of heat released (J), m = mass of solute (grams), c = specific heat capacity of water (
`4.18`×
`10^(3)`
`JKg^(-1)K^(-1)`), and ∆T=change in temperature.

∴ q = (0.1277)(4.18×
`10^(3)`)(2.98)

= 1590.68 J

= 1.59kJ

As enthalpy change is the energy required to change one mole of substance. Therefore, ∆H = q/n, where q = quantity of energy released/absorbed (which we already calculated), and n = number of moles of solvent (which we need as HCl).

Thus, n = cV = (0.500)(0.2) [remembering to change units] = 0.1 mol,

∴ ∆H = - 15.9 kJ/mol HCl. Remember that if the reaction is exothermic (releases heat), then enthalpy change is negative. If endothermic (absorbs heat), then enthalpy change is positive.

I'm not sure where you got -24.9 kJ/mol, I've double checked my work, I'm sure I'm correct.