Methanol is burned in a bomb calorimeter. Liquid water is formed as a product. If 34.0 g of methanol reacts, what is the expected temperature change in a calorimeter with a heat capacity of 6.75 kJ/C

Md Ariful Islam

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2CH3OH+ 3O2 = 2CO2 + 4H2O ∆H = 4(-285.8 kJ/mol) + 2(-393.510 kJ/mol) ndash; (2(-201.0 kJ/mol)) = -1528.22 kJ/mol n(CH3OH) = 34 g / 32g/mol = 1.0625mol Q = ∆H*n Q = 1528.22 kJ/mol* 1.0625mol = 1623.73 kJ ∆t = Q/C ∆t = 1623.73 kJ / 6.75kJ/C = 240.5 C
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