Both questions 12 & 13 When 0.187 g of benzene, C_6H_6, is burned in a bomb calo

Question

Both questions 12 & 13

When 0.187 g of benzene, C_6H_6, is burned in a bomb calorimeter, the surrounding water bath rises in temperature by 7.48 degree C. Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate combustion energies Delta E for benzene in both kilojoules per gram and kilojoules per mole. When a solution containing 8.00 g of NaOH in 50.0 g of water at 25.0 degree C is added to a solution of 8.00 g of HC1 in 250.0 g of water at 25.0 degree C in a calorimeter, the temperature of the solution increases to 33.5 degree C. Assuming that the specific hear of the solution is 4.18 J/(g. degree C) and that the calorimeter absorbs a negligible amount of heat, calculate Delta H [in kilojoules) for the reaction NaOH (aq) + HC1 (aq) rightarrow NaCl (aq) + H_20 (1) When the experiment is repeated using a solution of 10.0 g of HC1 in 248.0 g of water, the same temperature increase in observed. Explain.

Explanation / Answer

Change in internal energy = mass* specific heat* temperature difference

= 250gm* 4.184j/gm.deg.c* 7.48=7824 joules

Change in internal energy/gm= 7824/0.187=41840 J/gm=41.840 Kj/gm

Mass of benzene = 0.187gm, moles of Benzene = mass/molar mass = 0.187/78=0.00239

Change in internal energy per moles = 7824/0.00239 =3263520 joules/mole= 3263.52 Kj/mole

2.

2.Total mass = 8+50+8+250 = 316 gm

Temperature rise = 33.5-25= 8.5 deg.c

Enthalpy change = mass* specific heat* temperature difference= 316*4.184*8.5 joules=11238 joules=11.238 Kj

Moles of NaOH in 8 gm =8/40 =0.2 moles of HCl in 8gm = 8/36.5=0.22

As per the reaction, 1 mole of NaOH has to react with 1 mole of HCl . So HCl is excces here. So 0.22-0.2 = 0.2 moles og HCl remains unreacted.

It is true when 10 moles of HCl is added to 250 gm of water. Excess HCl remains and does not participate in the reaction of neutralization and not increase in enthalpy change.

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