Question

5 a) A Student carried out the following reaction in a coffe cup calorime cacos+ 2HCl Co+tho + Cachy She used 4.2g of reactant and 170.2g of acid. The initial Temprature of the Soyston was 24.0c and the final temprature was 26.0°C. Assuming the specifte heat capacity of the solution is 4.18019.c. calculate the total energy released or absored by the system in J. Poudel to nearest while number. b) Ethanol and carbon dioxide are Product in the fermenution reaction of glucose. The enthalples of combustion of glucon (C64 %) and ethanol (C₂H5OH) are 2813 kJ/mol and -1369 kJ/mol. calculate the molar enthalpy of the fermenation reaction given by the equation belows C6H₂O 9 2ę HOH (1) + 2 Cą (g) dont forgent put tor- stjn. C) is it exothermic or Endothermic? actional complexe d) use bond energy calculate entalpy chage pos: the (9) + f2 (9) 2Hfcg) Product reactent

Answer 1

CaCO3 +2 HC1 – CO2 + H2O + CaCl2 Calculate the total energy that was released or absorbed by the system (0) Q = m solution Csolution (T; – T;) m solution=solution mass msolution = 4.2 g CaCO3 + 170.2 g HCl = 174.4 g Csolution = specific heat of solution=4.18 J/g°C Ty=final temperature= 26°C T;=final temperature= 24°C Q = 174.4 9 x 4.18 - OC (26°C – 24°C) Q = 1458)

Write the combustion reaction for C6H12O6 and CH OH C6H12O6 (s) + 6 02 (g) →6 CO2 (g) + 6 H20 (1) AH° = -28132 C2H5OH (1) + 3 02 (g) → 2 CO2 (g) + 3 H20 (1) AH° = -1369 Multiply the ethanol combustion by 2 to get 2 mol of ethanol 2 x (C2H5OH (1) + 3 O2 (g) → 2 CO2 (g) + 3 H20 (1) AH° = -1369 2 C2H5OH (1) + 6 O2(g) → 4 CO2(g) + 6H2O (1) AH = -2738 Invert the ethanol combustion to get the ethanol as product 4 CO2 (g) + 6H20 (1) 2 C2H5OH (1) +6 02 (9) AH° = 2738" mol Add both combustion reactions to get the fermentation reaction C6H12O6 (s) +6 O2(g) → 26 CO2 (g) +6H20 (1) AH° = -2813 */ 4 CO2 (8) + 6H20 4)+ 2 C2H5OH (1) +6 02 (9) AH° = 2738 C6H1206 (s) – 2 CO2 (g) + 2 CH3OH (1) Calculate the AH, X own con=-2813 M2 + 2738 mol

SHun =-75 ble mol We know, AH,sn = {n,;, -En,9H;, Ip and n=stoichiometric coefficients of products and reactants. AHp and AH; r = standard formation enthalpy of products and reactants. From the graph, we can see that the product energy is higher than the reactants energy so that, AHxn is positive. If AHrxn is positive the reaction is endothermic.

H2(g) + F2 (g) → 2 HF (g) AH xn => nEbroken bond - m Eformed bond n and m= number of bonds that are broken or formed. Ebroken bond and Eformed bond= bond energy Table value mol Ex-x = 567 m Ex-p = 155 km mol k] Ehh = 436 mol AH x = (1 x EH-H + 1 x Ef-p) - (2 xEH-P) aktion = (1 + 436 mint 1 15 h ) - (2.4567 kl AH, = 591 -1134 – mol mol ? cn = -543 med mol