Question

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose by 4.20 Celsius. If the heat capacity of the bomb plus water was 10.4 kJ/degree Celsius, calculate the molar heat of combustion of methanol.

I've already completed the work and would like to check whether my answer is correct or not. I have this feelings that I messed up the significant figures when calculating the final result, so please let me know where I went wrong. My answer was -7.282 x 10² kJ

Answer 1

Answer:

A quantity of 1.922g of methanol (CH3OH) was burned in a constant -volume bomb calorimeter. consequently, the temperature of the water rose by 4.20degree centigrade. if the heat capacity of the bomb plus water was 10.4kj/degree centigrade, calculate the molar heat of combustion of methanol.

Answer 2

Answer:

The methanol's molar heat of combustion, $\Delta H$, calculated is $-728kJ/mol$.

Explanation:

Given,

The mass of methanol ( $CH_{3}OH$ ), m = $1.922g$

The water's temperature change, $\Delta T$ = $4.20\textdegree C$

The heat capacity of the bomb + water, C = $10.4kJ/\textdegree C$

The methanol's molar heat of combustion, $\Delta H$ =?

Firstly, we have to calculate the heat required for the burning of methanol ( q ) by using the formula given below:

• $q = C\Delta T$
• q = $10.4\times 4.20$
• q = $43.68kJ$

Now, the molar heat of combustion ( $\Delta H$ ) is given by:

• $\Delta H = -\frac{q}{n}$      -------equation (1)

The negative sign indicates that the reaction is exothermic.

Here, n = The number of moles of methanol

As we know, the molar mass of methanol = $32g/mol$

Therefore,

• The number of moles of methanol = $\frac{Given mass}{Molar mass}$ = $\frac{1.922}{32}$ = $0.06mol$

After putting the calculated values of moles and heat in equation (1), we get:

• $\Delta H = -\frac{43.68}{0.06}$
• $\Delta H = -728kJ/mol$

Hence, the methanol's molar heat of combustion, $\Delta H$ = $-728kJ/mol$.