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Calculating Heat of Combustion for Benzene

Calculate the heat given off when 1 mol of benzene undergoes combustion in a bomb calorimeter using temperature change and heat capacity. Formula: C6H6(l) + 15/2 O2(g) -> 6CO2(g) + 3H2O(l). Follow the step-by-step process to determine the heat transfer.

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Calculating Heat of Combustion for Benzene

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  1. A 3.51 g sample of benzene, C6H6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00oC and rose to 37.18oC. The heat capacity of the calorimeter and contents was 12.05 kJ/oC. Find q (heat given off) for the combustion of 1 mol of benzene.

  2. A 3.51 g sample of benzene, C6H6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00oC and rose to 37.18oC. the heat capacity of the calorimeter and contents was 12.05 kJ/oC. Find q for the combustion of 1 mol of benzene. C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) 3.51 g What units do we desire? T = 12.18oC

  3. A 3.51 g sample of benzene, C6H6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00oC and rose to 37.18oC. the heat capacity of the calorimeter and contents was 12.05 kJ/oC. Find q for the combustion of 1 mol of benzene. C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) 3.51 g T = 12.18oC

  4. If we could find the number of kJ and the number of moles of benzene, we could divide the kJ by the moles to get our answer!

  5. A 3.51 g sample of benzene, C6H6, was burned in a bomb calorimeter in an excess of oxygen. The initial temperature was 25.00oC and rose to 37.18oC. the heat capacity of the calorimeter and contents was 12.05 kJ/oC. Find q for 1 mol of benzene. C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) 3.51 g T = 12.18oC 12.18 oC 146.8 = kJ 3.51 g = mol BZ 0.04497

  6. Find q for 1 mol of benzene. C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) 3.51 g T = 12.18oC 12.18 oC 146.8 = kJ 3.51 g = mol BZ 0.04497 And now?

  7. Find q for 1 mol of benzene. C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) 3.51 g T = 12.18oC 12.18 oC 146.8 = kJ 3.51 g = mol BZ 0.04497 146.8 kJ 3264 0.04497 mol Bz

  8. Find q for 1 mol of benzene. C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l) 3.51 g T = 12.18oC 12.18 oC = 3264 3.51 g

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