When a 0.150 g sample of Ca metal is combined with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter, leading to a temperature increase from 23.5 °C to 25.4 °C, the enthalpy of the reaction is -2.12 × 10⁵ J/mol.
A 0.150 g sample of Ca metal is combined with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter. If we assume the density of the solution to be equal to the density of the water (1 g/mL), the mass of the solution is:
[tex]100.0 mL \times \frac{1g}{mL} = 100.0g[/tex]
We can find the heat absorbed by the solution (Qs) using the following equation.
[tex]Qs = c \times m \times \Delta T = \frac{4.18J}{g.\° C} \times 100.0g \times (25.4\° C - 23.5 \° C) = 794 J[/tex]
where,
- c: specific heat capacity of the solution (we will assume it is equal to that of water)
- m: mass of the solution
- ΔT: change in the temperature
According to the law of conservation of energy, the sum of the heat absorbed by the solution (Qs) and the heat released by the reaction (Qr) is zero.
[tex]Qs + Qr = 0\\\\Qr = -Qs = -794 J[/tex]
The molar mass of Ca is 40.08 g/mol. The moles (n) corresponding to 0.150 g of Ca are:
[tex]0.150 g \times \frac{1mol}{40.08g} = 3.74 \times 10^{-3} mol[/tex]
Finally, we can calculate the enthalpy of the reaction (ΔHrxn) using the following expression.
[tex]\Delta H_{rxn} = \frac{Qr}{n} = \frac{-794 J }{3.74 \times 10^{-3} mol} = -2.12 \times 10^{5} J/mol[/tex]
When a 0.150 g sample of Ca metal is combined with enough HCl to make 100.0 mL of solution in a coffee-cup calorimeter, leading to a temperature increase from 23.5 °C to 25.4 °C, the enthalpy of the reaction is -2.12 × 10⁵ J/mol.
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