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The aim of this experiment is to calculate the enthalpy change in various alcohols.

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Introduction

CHEMISTRY 3: Energy, rate and extent of reaction AIM The aim of this experiment is to calculate the enthalpy change in various alcohols. HYPOTHESIS Energy is produced when a fuel is burned. The fuel can be used to heat water in a calorie meter. In order to calculate the energy produced by the fuel, the mass, specific heat capacity and temperature of the water need to be known. The enthalpy change can then be calculated using the following formula: ?H=mc?t Where ?H=Enthalpy change, m=mass, c=specific heat capacity, and ?t=change in temperature. Specific heat capacity is the energy needed by one gram of a substance to raise its temperature by 1�K. The specific heat capacity of water is a standard of 4.2J/gk. In this experiment, the fuel to be used is alcohol. ...read more.

Middle

content onto base of stand * Place calorie meter onto clamp and lower so that the base of the calorie meter is approximately 2cm above the spirit lamp. DO NOT adjust this height when repeating the procedure * Measure 50cm� of distilled water * Pour into calorie meter * Put cork lid onto calorie meter, and place thermometer through the central hole in the lid. Take care to make sure that the thermometer is not touching the base of the calorie meter * Wrap a loose layer of aluminum foil around the clamp stand to cover the calorie meter and spirit lamp, leaving a small gap at the base * Record the start temperature of the water * Light spirit lamp and record temperature of water at 30 second intervals for three minutes * After three minutes, extinguish flame on spirit lamp * ...read more.

Conclusion

CALCULATIONS AND OBSERVATIONS Methanol: CH3OH ?H=mc?t ?T=49-22 =27�C ?H=50x4.2x27 ?H=5,670j/g 5670/1000 =5.67 ?H=5.674j/kg Propanol: C3H7OH ?H=mc?t ?T=50-20 =30�C ?H=50x4.2x30 ?H=6,300j/g 6300/1000 =6.3 ?H=6.3j/kg Pentanol: C5H11OH ?H=mc?t ?T=48.5-23 =25.5�C ?H=50x4.2x25.5 ?H=5,355j/g 5355/1000 =5.355 ?H=5.355j/kg Octanol: C8H17OH ?H=mc?t ?T=52-21 =31�C ?H=50x4.2x31 ?H=6,510j/g 6510/1000 =6.51 ?H=6.51j/kg Butanol: C4H9OH ?H=mc?t ?T=45.5-22.5 =23�C ?H=50x4.2x23 ?H=4,830j/g 4830/1000 =4.83 ?H=4.83j/kg Hexanol: C6H13OH ?H=mc?t ?T=47-22.5 =24.5�C ?H=50x4.2x24.5 ?H=5,145j/g 5145/1000 =5.145 ?H=5.145j/kg Ethanol: C2H5OH ?H=mc?t ?T=57-24 =33�C ?H=50x4.2x33 ?H=6,930j/g 6930/1000 =6.93 ?H=6.93j/kg CONCLUSION The principle is that the more carbon atoms that are present in the alcohol molecule, the more energy is given out. As shown in this experiment, Butanol is the worst performer as it was the least energy efficient. Ethanol is the best for combustion, as it is the most energy efficient. To improve this experiment, an increased range of alcohol could be used in order to support this conclusion. ...read more.

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