Answer:
a) Q = 2047.8 J (ΔH is negative because it's an exothermic reaction)
b) ΔH = -12.7 kJ /mol
Explanation:
Step 1: Data given
Molar mass of X = 78.2 g/mol
In a constant-pressure calorimeter, 12.6 g of X is dissolved in 337 g of water at 23.00 °C.
The temperature rise to 24.40 °C
The specific heat of the solution = 4.184 J/g°C
Step 2: Calculate the total mass
Total mass of the solution is given by
Total mass = 12.6 grams + 337 grams = 349.6 grams
Step 3: Calculate heat
Q = m*c*ΔT
⇒ m = the total mass = 349.6 grams
⇒ c = the specific heat of solution = 4.184 J/g°C
⇒ ΔT = The change of temperature = T2 - T1 = 24.40 - 23.00 = 1.40 °C
Q = 2047.8 J (ΔH is negative because it's an exothermic reaction)
What is the enthalpy of the reaction?
Calculate number of moles = mass/ molar mass
Moles X = 12.6 grams / 78.2 g/mol
Moles X = 0.161 moles
ΔH = -2047.8 J / 0.161 moles
ΔH = -12719.3 J/mol = -12.7 kJ /mol
A solution prepared by dissolving 12.6 g of X in 337 g of water, whose temperature increases from 23.00 °C to 24.00 °C, absorbs 2.05 × 10³ J of heat. The enthalpy of the reaction is -12.7 kJ/mol.
We have a solution prepared by dissolving 12.6 g of X (solute) in 337 g of water (solvent). The mass of the solution (m) is:
[tex]m = 12.6g + 337 g = 350. g[/tex]
The temperature of the solution increases from 23.00 °C to 24.40 °C. Assuming that the solution has the same specific heat as water (c = 4.184 J/(g·°C)), we can calculate the heat absorbed (Q) by the solution using the following expression.
[tex]Q = c \times m \times \Delta T = \frac{4.184J}{g.\° C} \times 350. g \times (24.40 \° C - 23.00 \° C) = 2.05 \times 10^{3} J[/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 is zero.
[tex]Qsol + Qreaction = 0\\Qreaction = -Qsol = -2.05 \times 10^{3} J[/tex]
The dissolution of 12.6 g of X (molar mass 78.2 g/mol) leads to the release of 2.05 × 10³ J (hence the negative sign). The enthalpy of the reaction is
[tex]\Delta H^{0} = \frac{-2.05 \times 10^{3} J}{12.6g} \times \frac{78.2g}{1mol} = -1.27 \times 10^{4} J/mol = -12.7 kJ/mol[/tex]
A solution prepared by dissolving 12.6 g of X in 337 g of water, whose temperature increases from 23.00 °C to 24.00 °C, absorbs 2.05 × 10³ J of heat. The enthalpy of the reaction is -12.7 kJ/mol.
You can learn more about calorimetry here: https://brainly.com/question/16104165
