In a calorimetric experiment, 6.48 grams of lithium hydroxide was dissolved in water. The temp of the calorimeter rose from 25 celcius to 36.66 celcius. What is the molar change in enthalpy for this procces?

LiOH(s) -> Li(aq) + OH (aq)

(the heat capacity of the calorimeter and its contents is 547 J/C; molar mass of LiOH= 23.94g)

To calculate the molar change in enthalpy (ΔH) for this process, we can use the equation:

q = m x C x ΔT

Where:
q is the heat absorbed or released during the reaction
m is the mass of the substance being dissolved (in this case, lithium hydroxide)
C is the heat capacity of the calorimeter and its contents
ΔT is the change in temperature

Step 1: Calculate the heat absorbed or released (q) during the reaction.

First, we need to calculate the temperature change (ΔT):
ΔT = final temperature - initial temperature
ΔT = 36.66°C - 25°C
ΔT = 11.66°C

Next, we can calculate q using the equation:
q = m x C x ΔT

m = 6.48 g (mass of lithium hydroxide)
C = 547 J/°C (heat capacity of the calorimeter and its contents)

q = 6.48 g x 547 J/°C x 11.66 °C
q = 42363.27 J

Step 2: Calculate the moles of LiOH dissolved.

We can use the molar mass of LiOH to convert grams to moles:
Molar mass of LiOH = 23.94 g/mol

moles of LiOH = 6.48 g / 23.94 g/mol
moles of LiOH = 0.27 mol

Step 3: Calculate the molar change in enthalpy (ΔH).

ΔH = q / moles of LiOH dissolved
ΔH = 42363.27 J / 0.27 mol

ΔH = 156900 J/mol

Therefore, the molar change in enthalpy (ΔH) for this process is 156900 J/mol.

To find the molar change in enthalpy, we need to use the equation:

q = m * C * ΔT

where:
- q is the heat absorbed or released in the reaction (in Joules, J)
- m is the mass of the substance (in grams, g)
- C is the heat capacity of the calorimeter and its contents (in J/°C)
- ΔT is the change in temperature (in °C)

First, let's calculate the heat absorbed or released by the reaction:

m = 6.48 g (given)

ΔT = 36.66 °C - 25 °C = 11.66 °C

q = m * C * ΔT
q = 6.48 g * 547 J/°C * 11.66 °C
q ≈ 4209.528 J

Next, we need to calculate the number of moles of LiOH used in the reaction:

molar mass of LiOH = 23.94 g/mol (given)

moles of LiOH = mass / molar mass
moles of LiOH = 6.48 g / 23.94 g/mol
moles of LiOH ≈ 0.2705 mol

Now, we can calculate the molar change in enthalpy using the formula:

ΔH = q / moles of LiOH

ΔH = 4209.528 J / 0.2705 mol
ΔH ≈ 15565.05 J/mol

So, the molar change in enthalpy for this process is approximately 15565.05 J/mol.

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