S1237
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Can someone explain this question in steps please
In an experiment, 0.750 g of benzene (C6H6) were completely burned in air. The heat evolved raised the temperature of 200 g of water by 43.7°C. Use this data to calculate the enthalpy of combustion of benzene
(the specific heat capacity of water is 4.18 J g-1 K-1)
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Aethomson
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Surely it’s just energy = mass x temp change x specific heat capacity. (Using the values for water as that is what is being heated) then the energy that was used to heat the water is the same as the energy released by the benzene.

This is what I think it’s been a while since I’ve done this topic!
Last edited by Aethomson; 6 months ago
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username5567480
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As @Aethomson said, you use q = mass x specific heat capacity x temperature change. So it would be 200 x 4.18 x 43.7 = 36533.2 J (36.5332 kJ)

Then work out how many moles of benzene is produced by using mass/mr. They’ve given you the mass of benzene which is 0.750 so you divide that by the mr of benzene. So it would be 0.750/78 = 0.0096...

Then work out how much heat is produced by 1 mole. 36.5332/0.0096... = - 3799.45 kJ mol-1 (you add the negative sign because combustion is an exothermic reaction).
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S1237
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(Original post by Aethomson)
Surely it’s just energy = mass x temp change x specific heat capacity. (Using the values for water as that is what is being heated) then the energy that was used to heat the water is the same as the energy released by the benzene.

This is what I think it’s been a while since I’ve done this topic!
Thank you
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S1237
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(Original post by DSKE)
As @Aethomson said, you use q = mass x specific heat capacity x temperature change. So it would be 200 x 4.18 x 43.7 = 36533.2 J (36.5332 kJ)

Then work out how many moles of benzene is produced by using mass/mr. They’ve given you the mass of benzene which is 0.750 so you divide that by the mr of benzene. So it would be 0.750/78 = 0.0096...

Then work out how much heat is produced by 1 mole. 36.5332/0.0096... = - 3799.45 kJ mol-1 (you add the negative sign because combustion is an exothermic reaction).
Thank you so much
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