Answer:
The enthalpy change in the the reaction is -47.014 kJ/mol.
Explanation:
[tex]X(s)+H_2O(l)\rightarrow X(aq)[/tex]
Volume of water in calorimeter = 22.0 mL
Density of water = 1.00 g/mL
Mass of the water in calorimeter = m
[tex]m=1.00 g/mL\times 22.0 mL=22 g[/tex]
Mass of substance X = 2.50 g
Mass of the solution = M = 2.50 g + 22 g = 24.50 g
Heat released during the reaction be Q
Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C
Specific heat of the solution is equal to that of water :
c = 4.18J/(g°C)
[tex]Q=Mc\times \Delta T[/tex]
[tex]Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ[/tex]
Heat released during the reaction is equal to the heat absorbed by the water or solution.
Heat released during the reaction =-1.433 kJ
Moles of substance X= [tex]\frac{2.50 g}{82.0 g/mol}=0.03048 mol[/tex]
The enthalpy change, ΔH, for this reaction per mole of X:
[tex]\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol[/tex]
2.50 g of X dissolves in a calorimeter containing 22.0 mL of water, causing the temperature to increase from 14.0 °C to 28.0 °C. The enthalpy change of the reaction is -46.9 kJ/mol.
First, we will convert 22.0 mL of water to grams using its density (1.00 g/mL).
[tex]22.0 mL \times \frac{1.00g}{mL} = 22.0 g[/tex]
The solution contains 22.0 g of water and 2.50 g of X. The mass of the solution is:
[tex]m = 22.0 g + 2.50 g = 24.5 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 24.5 g \times (28.0 \° C - 14.0 \° C) = 1.43 \times 10^{3} J = 1.43 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the solution and the heat released by the reaction (Qr) is zero.
[tex]Qs + Qr = 0\\\\Qr = -Qs = -1.43 kJ[/tex]
Then, we will convert 2.50 g of X to moles using its molar mass (82.0 g/mol).
[tex]n = 2.50 g \times \frac{1mol}{82.0g} = 0.0305 mol[/tex]
Finally, we will calculate the enthalpy change, ΔH, for this reaction per mole of X using the following expression.
[tex]\Delta H = \frac{Qr}{n} = \frac{-1.43 kJ}{0.0305mol} = -46.9 kJ/mol[/tex]
2.50 g of X dissolves in a calorimeter containing 22.0 mL of water, causing the temperature to increase from 14.0 °C to 28.0 °C. The enthalpy change of the reaction is -46.9 kJ/mol.
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