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Unformatted text preview: consumed by the reaction. 52) a) An increase in temperature would increase the energy of the system closer to the activation energy and speed up the reaction. b) The effect of a temperature increase on the rate constant of the reverse reaction would also be an increase in the rate of reaction. Increasing the temperature increases the rate of the reverse reaction to a greater extent than it does the rate of the forward reaction. 53) a) The heat of the reaction can be expressed as H=(H f products ) – (H f reactants ) = (-110.0 kJ/mol + 34.0 kJ/mol) - (-393.5 kJ/mol + 90.4 kJ/ mol) =227.1 kJ/mol b) Activation energy reverse = Activation energy forward- H forward =(135 kJ/mol – 227.1 kJ/mol) = -92.1 kJ/mol c) 54) Experiment #4 can be found on the next pages....
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- Spring '09
- Chemical reaction, 34.0 kJ, 90.4 kJ, 92.1 kJ, 110.0 kJ, 135 kJ