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At 273 K the enthalpy of a certain reaction is -20.75 kJ. If heat capacities of the reactants and products are the same, then what will be the enthalpy of the same reaction at 373 K?


(a) -20.75 kJ
(b) -2075 kJ
(c) 0
(d) -20.75 $\times \frac{373}{273}$ kJ

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Answer: -20.75 kJ
By Kirchoff's equation $\frac{\Delta H_2 - \Delta H_1}{T_2 - T_1} = \Delta C_P$
Since, ($C_P$) reactant = ($C_P$) Products , $\Delta C_P = 0$
Hence, $\Delta H_2 - \Delta H_1 = 0 \Longrightarrow \Delta H_2 =\Delta H_1 = -20.75$ kJ
answered Feb 27, 2014 by mosymeow_1
 

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