A calorimeter contains 21.0 mL of water at 12.0 degress Celsius. When 2.40g of X (a substance with a molar mass of 50.0 g/mol ) is added, it dissolves via the reaction
X(s)+H2O(l) --> X (aq)
and the temperature of the solution increases to 29.5 degress celsius.
Calculate the enthalpy change, H , for this reaction per mole of X .
Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 J/(g* degress celsius and 1.00 g/mL ] and that no heat is lost to the calorimeter itself, nor to the surroundings.
Express your answer numerically in kilojoules per mole.
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-30.2 kJ/mol
To calculate the enthalpy change per mole of X, we can use the equation:
q = m * c * ΔT
where:
q is the heat absorbed or released by the reaction
m is the mass of the water (density * volume)
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's calculate the mass of water used:
density of water = 1.00 g/mL
volume of water = 21.0 mL
mass of water = density * volume
mass of water = 1.00 g/mL * 21.0 mL
mass of water = 21.0 g
Next, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 29.5°C - 12.0°C
ΔT = 17.5°C
Now, we can calculate the heat absorbed or released by the reaction:
q = mass * specific heat capacity * ΔT
q = 21.0 g * 4.18 J/(g * °C) * 17.5°C
q = 1551.15 J
To convert Joules to kilojoules, divide q by 1000:
q = 1551.15 J / 1000
q = 1.55115 kJ
Finally, to calculate the enthalpy change per mole of X, we need to use the molar mass of X.
molar mass of X = 50.0 g/mol
The enthalpy change per mole is given by:
H = q / moles of X
H = 1.55115 kJ / (2.40 g / 50.0 g/mol)
H = 1.55115 kJ / (0.0480 mol)
H = 32.3219 kJ/mol
Therefore, the enthalpy change for this reaction per mole of X is approximately 32.32 kJ/mol.