Here is the question:
A sample of ethanol, C2H5OH, weighing 2.84g was burned in an excess of oxygen in a bomb calorimeter. THe temperature of the calorimeter rose from 24C to 33.73C. If the heat capacity of the calorimeter and contents was 9.63 kj/C. What is the calue of q for burnign 1 mol of ethanol at constant volume and 25.00C?
C2H5OH(l)+3O2(g)-->2CO2(g)+3H2O(l)
Here is my Homework:
q=ms(T final - T initial)
q=2.84x9.63kj/C x (33.73C-25C)= 238.76
But 238.76 is not the correct answer, the book states that it is -00136kJ
Can anyone tell me what I am doing incorrect?
You figured the q for 2.84g. Now figure it for a mole of ethanol.
Use a proportion.
2.48/46.068g/mol ethanol=.06165
.06165x9.63kj/C x (33.73C-25C)= 5.183 ??
still not the -00136kJ I am looking for??
To solve this problem, you need to consider the stoichiometry of the reaction and apply the concept of molar heat of combustion or molar enthalpy of combustion.
First, let's calculate the moles of ethanol burned:
Given:
Mass of ethanol = 2.84 g
Molar mass of ethanol = 46.07 g/mol
Moles of ethanol = Mass of ethanol / Molar mass of ethanol
= 2.84 g / 46.07 g/mol ≈ 0.0617 mol
Now, we need to calculate the heat produced per mole of ethanol burned.
In the balanced equation, 1 mole of ethanol is burned, and it produces q amount of heat. This means that the heat produced per mole of ethanol burned is the same as q. Therefore:
Heat per mole of ethanol burned, q = 238.76 kJ / 0.0617 mol ≈ 3873 kJ/mol
Please note that in your earlier calculation for q, you have correctly multiplied the mass, heat capacity, and temperature difference. However, the result you obtained is not the value for q per mole, but rather the total heat generated for the given mass of ethanol.
The expected result of -0.136 kJ is most likely due to a sign error. The negative sign indicates that the reaction is exothermic, meaning it releases heat. In this case, it suggests that your answer should be negative.
So, the correct calculation should yield q = -3873 kJ/mol, where the negative sign indicates the exothermic nature of the reaction.
Remember to always double-check the signs when performing calculations involving heat changes.