The change in reaction enthalpy is 49305.02 J / mol
C = 4.18
delta T = 24.25 - 23.16
= 1.09 mol
= 0.070L * 0.185M
= 0.01295 m
= d * v
= 1.00 * (0.070 + 0.070)
= 0.14g
= q / mol HCl
= [mC (deltaT)] / mol HCl
= (0.14 * 4.18 * 1.09) / (0.01295)
= 49.3 kJ
Molarity of solution = Moles of solute x 1000 / Volume of solution (in mL)
Molarity of HCl solution = 0.185 M
The volume of the solution = 70.0 mL
0.185 = Mole HCl x 1000 / 70.0
Mole HCl = 0.185 x 70.0 / 1000 = 0.01295mol
\ Delta H_ (rxn) = (q) (n)
Where,
q = amount of heat absorbed = 638.5 J
n = number of moles = 0.01295 moles
\ Delta H_ (rxn) = change in enthalpy of reaction
\ Delta H_ (rxn) = 638.5J / 0.01295mol = 49305.02J / mol
Therefore, the change in reaction enthalpy is 49305.02 J / mol
Enthalpy (H) is the chemical energy contained in a system. The enthalpy of a system cannot be measured, which can be measured is the change in enthalpy (ΔH) that accompanies the change in the system.
The standard enthalpy of formation is the heat absorbed or released in the formation of one mole of substances from the elements measured at 250C and 1 atmospheric pressure. A standard decomposition enthalpy is a heat absorbed or released in the decomposition of one mole of substances into its elements.
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Grade: Middle School
Subject: Chemistry
Keyword: Molarity, HCl, enthalpy
When 50.0 mL of 0.100 M AgNO₃ is combined with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, leading to a temperature change from 23.40 °C to 24.21 °C, the enthalpy change for the reaction is -67.8 kJ/mol.
Let's consider the following reaction that takes place in a coffee-cup calorimeter.
AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)
50.0 mL of 0.100 M AgNO₃ is combined with 50.0 mL of 0.100 M HCl. The reacting moles of each reactant are:
[tex]n = 0.0500 L \times 0.100mol/L = 5.00 \times 10^{3} mol[/tex]
The volume of the final solution will be the sum of the volumes of each solution. Considering the density of the solution is 1.00 g/mL, the mass of the solution is:
[tex](50.0mL+50.0mL) \times 1.00g/mL = 100.0 g[/tex]
We can calculate the heat absorbed by the solution (Qs) using the following expression.
[tex]Qs = c \times m \times \Delta T = \frac{4.18J}{g.\° C } \times 100.0g \times (24.21\° C-23.40\° C) = 339 J[/tex]
where,
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 = -339 J[/tex]
We can calculate the enthalpy change for the reaction (ΔHrxn) using the following expression.
[tex]\Delta Hrxn = \frac{Qr}{n} = \frac{-339J}{5.00 \times 10^{-3}mol } \times \frac{1kJ}{1000J} = -67.8 kJ/mol[/tex]
When 50.0 mL of 0.100 M AgNO₃ is combined with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, leading to a temperature change from 23.40 °C to 24.21 °C, the enthalpy change for the reaction is -67.8 kJ/mol.
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