A student used the apparatus shown to determine the molar heat of combustion of ethanol - HSC - SSCE Chemistry - Question 4 - 2006 - Paper 1

Change in temperature = Final temperature of water - Initial temperature of water Change in temperature = 45.5°C - 25.0°C = 20.5°C

Using the formula:

$q = mc\Delta T$

where:

• m = mass of water = 300 g = 0.300 kg
• c = specific heat capacity of water = 4.18 kJ kg⁻¹°C⁻¹
• ( \Delta T = 20.5°C )

Energy absorbed by water:

$q = 0.300 \times 4.18 \times 20.5 = 25.743 \text{ kJ}$

To find the molar heat of combustion (ΔH), we can use the formula:

$\Delta H = \frac{q}{n}$

where:

• q = energy released = -25.743 kJ (negative because energy is released)
• n = number of moles of ethanol burned = ( \frac{1.15 ext{ g}}{46.07 ext{ g/mol}} \approx 0.0250 ext{ mol} )

Now, substituting these values:

$\Delta H = \frac{-25.743}{0.0250} \approx -1029.72 \text{ kJ/mol}$

Rounding gives approximately -1030 kJ mol⁻¹. Therefore, the molar heat of combustion is:

C