College of Arts & Seiences Department of Chemistry & Earth Sciences sample of th
ID: 551366 • Letter: C
Question
College of Arts & Seiences Department of Chemistry & Earth Sciences sample of the sugar D-ribose (CsHio0s) of mass 0.727 g was placed in a bomb calorimeter and .910 K. In a separate experiment then ignited in the presence of excess oxygen. The temperature rose by 0 calorimeter, the combustion of 0.825 g of benzoie acid, for which the internal energy of combustion is-3251 J moH, gave a temperature rise of .940 K. Heat contribution of wire combustion i: neglected. a) Write balanced chemical equation for the combustion reaction of D-Ribose b) Calculate the internal energy of combustion of D-ribose. c) Calculate the enthalpy of combustion of D-ribose. (Hint: no change in the number of moles of gas) 393.5 kJ/mol: n obtained frorn the experiment. calculate 11 r of D ribose . All' CO2(g) ofD ribose reportedin the literature's 1267kJ/mol calculate e) I the l the % error of the experimen measurement. w your workExplanation / Answer
C5H10O5+ 5O2----->5CO2 (g)+ 5H2O (l)- balanceed chemical reaction
change in internal energy= 3251 Kj/mole, moles of benzoic acid = mass /molar mass =0.825/122=0.006762
change in internal energy= 3251*0.006762 =21.96 KJ
heat capacity of calorimter* temperature rise
21.96= heat capacity of calorimter*1.940
Heat capacity of calorimeter= 21.96/1.940 KJ/K=11.32KJ/K
internal energy of combustion of D- ribose = - heat capacity of calorimeter* temperature rise = -11.32*0.910KJ=-10.3KJ
moles of ribose (C5H10O5)= mass of ribose/ molar mass of ribose , molar mass of ribose = 5*12+10*1+16*5=150 g/mole
moles of ribose = 0.727/150 =0.004847
internal energy change/moles = -10.3/0.004847=2125
H= U+PV
dH= dU+d(PV) =deltaU+deltan*RT , deltan= change in no of moles of gases = 5+-5= 0
deltaH= -2178= deltaU
5CO2(g)+ 5H2O(l) ------>C5H10O5+5O2, deltaH= 2125 KJ (1)
5C(s)+5O2(g)-------->5CO2(g), deltaH=-5*393.5 Kj (2)
5H2(g)+2.5O2(g)------->5H2O(l), deltaH =-5*285.8 (3)
Eq.1+Eq.2+Eq.3, 5C+ 5H2(g)+2.5O2 ------>C5H10O5 , deltaH= 2125-1967.5-1429 = -1269.5 KJ
enthalpy of formation (literature value)= -1267 Kj/mole
% difference = 100*(-1267+1269.5)/(-1267)=-0.197%