The combustion of methanol is shown by the following equation: 2CH3OH(l) + 3O2(g
ID: 858175 • Letter: T
Question
The combustion of methanol is shown by the following equation:
2CH3OH(l) + 3O2(g) --> 2CO2(g) + 4H2O(g)
i. Given the data which follows:
a. Find the heat of reaction for the equation above.
b. State the molar heat of combustion of methanol.
c. State whether the reaction is endothermic or exothermic.
C(s) + O2(g) --> CO2(g) + 393 kJ
H2(g) + 1/2O2(g) --> H2O(g) + 242 kJ
C(s) + 2H2(g) + 1/2O2(g) --> CH3OH(l) + 638 kJ
ii. What mass of water could be heated from 20.0 deg. C to 35.0 deg. C by the burning of 2.57 mol methanol? (C v H2O = 4.184 kJ/kg deg. C)
Explanation / Answer
The sign of the heat of formation matters. Using hess's law:
2(-238kJ) + 4(242kJ) + 2(393)= -1279=Heat of reaction
Heat of combustion of methanol is just the heat of reaction divided by the molar coefficient for methanol in the chemical equation so about -639.5kJ/mol . In acuality it is somewhere near -727kJ/mol. I may have calculated wrong.
Using the caluculated value:
2.57mol of methanol x -639.5kJ/mol = -1643.5kJ
1643.5kJ = (4.184)(15C)m m=26.2kg