A student is asked to calculate the amount of heat involved inchanging 100.00 g
ID: 681359 • Letter: A
Question
A student is asked to calculate the amount of heat involved inchanging 100.00 g of benzene gas (C6H6) at80.00oC to liquid benzene at 25.00oC. To do this you will need the information of the specific heat,boiling point, and the vaporization of benzene as listedbelow. In addition the following step-wise process must befollowed. a) Calculate the H for C6H6 (g,80.00oC) -----> C6H6 (g,25.00oC) b) Calculate the H for C6H6 (g,25.00oC)-----> C6H6 (l,25.00oC) c) Using Hess's law calulate the H for:C6H6 (g, 80.00oC)----->C6H6 (l, 25.00oC) Specific heat of Benzene is 1.72 J/g * oC BenzeneC6H6 mp(oC)5, Hfus9.84 bp(oC)80, Hvap30.8 The total answer is -48.9 kJ, I just dont quiteunderstand the process taken any and all help greatlyappreciated!!! A student is asked to calculate the amount of heat involved inchanging 100.00 g of benzene gas (C6H6) at80.00oC to liquid benzene at 25.00oC. To do this you will need the information of the specific heat,boiling point, and the vaporization of benzene as listedbelow. In addition the following step-wise process must befollowed. a) Calculate the H for C6H6 (g,80.00oC) -----> C6H6 (g,25.00oC) b) Calculate the H for C6H6 (g,25.00oC)-----> C6H6 (l,25.00oC) c) Using Hess's law calulate the H for:C6H6 (g, 80.00oC)----->C6H6 (l, 25.00oC) Specific heat of Benzene is 1.72 J/g * oC BenzeneC6H6 mp(oC)5, Hfus9.84 bp(oC)80, Hvap30.8 The total answer is -48.9 kJ, I just dont quiteunderstand the process taken any and all help greatlyappreciated!!!Explanation / Answer
given mass of benzene m = 100 molar mass of benzene =78 moles of benzene = 100/78 T =80-25=55 a)H for C6H6 (g,80.00oC) -----> C6H6 (g,25.00oC) is mCT = 100*1.72*55=9.46kJ ----------------------------------------------------------------- b)H for C6H6 (g,25.00oC)-----> C6H6 (l,25.00oC) ? given Hvap= 30.8kJ this is per mole , since 100/78moles of benzene was reacted, H for this phase-transformation =(100/78)*30.8kJ =39.48kJ ------------------------------------------------------------------------------------------ c)C6H6 (g,80.00oC)-----> C6H6 (l,25.00oC) this reaction can be achieved like this: C6H6 (g, 80.00oC) ----->C6H6 (g, 25.00oC) H = 9.46kJ C6H6 (g, 25.00oC)----->C6H6 (l,25.00oC) H =39.48kJ ------------------------------------------------------------------- adding these two reactions we get : C6H6 (g, 80.00oC)----->C6H6 (l, 25.00oC) and byhess's law H=H1+H2 => H = (9.46+39.48 )kJ so H for: C6H6 (g,80.00oC)-----> C6H6 (l,25.00oC) is 48.94kJ
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