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akhil samir killawala Grade: 11
        

RESPECTED SIR,


I HAVE THE FOLLOWING DOUBTS IN THERMO CHEMISTRY.


I WOULD ALSO LIKE TO KNOW WHICH IS THE BEST BOOK FOR THERMOCHEMISTY OTHER THAN PC BAHUDAR  AND R C MUKHERJEE?


1)  A PERSON TAKES 15 BREATHES PER MINUTE.THE VOL OF AIR INHALED IN EACH BREATH IS 448ML AND CONTAINS 21% OXYGEN BY VOL.  THE EXHALED AIR CONTAINS 16% OXYGEN BY VOL.  IF ALL THE OXYGEN IS USED IN COMBUSTION OF SUCROSE.


HOW MUCH HEAT IS EVOLVED.(THE HEAT OF COMBUSTION OF SUCROSE ID -600 kJ /MOL AND TEMP IS 300K )


2)THE STANDARD ENTHALPY OF SUCROSE IS -5645 KJ/MOL ..WHAT IS THE ADVANTAGE (IN KJ/MOL OF ENERGY EVOVLED AS HEAT) OF COMPLETE AEROBIC OXIDATION COMPARED TO ANAEROBIC HYDROLYSIS OF SUCROSE TO LACTIC ACID?


(HEAT OF FORMATION FOR LACTIC ACID , CO2 AND H20  IS - 


-694, -395 , 286 RESPECTIVELY)


 

7 years ago

Answers : (1)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


The enthalpy of reaction, dH, is the energy [heat (q) and work (d(P V))] released in a reaction. The thermochemical equation is usually written in the form: 2 H2(g, 1atm) + O2(g) = 2 H2O(l),       dH = -571.7 kJ This equation means that when 2 moles of H2 gas react with 1 mole of O2 gas, 571.7 kJ of energy is released (lost to the surrounding). Thus, if 1 mole of H2 and one half mole of O2 react, half of 571.7 kJ or 285.9 kJ of energy is released. H2 (g, 1atm) + 0.5 O2 = H2O(l),       dH = -285.9 kJ


On the other hand, if double amounts of reactants (i.e. 4 moles of H2 and 2 moles of O2) are used, twice amount of energy. (2*(-571.7) =) -1043.4 kJ is released.


If dH is positive, at least that much energy must be supplied to carry out the endothermic reaction.


The Standard Enthalpy of Reaction


For convenience in application, 1 atm for gas and 1.0 M for solutions were considered the "standard conditions", and data collected at standard conditions were called standard data such as standard enthalpy of reaction and standard enthalpy of formation. These values are condensed and summarized in handbooks for scientists and engineers in their applications.


Because temperature, pressure, and concentration of reactants and products affect the amount of measured energy, the scientific community has agree upon a temperature of 273 K and 1 atm as the standard temperature and and pressure (STP). However, standard enthalpies are often given for data collected at 298 K.


The most stable state at the standard condition is the standard state. The enthalpy of an element at its standard state is assigned 0 for reference.


For example, at 1 atm, graphite is the most stable state of carbon. The standard enthalpy of combustion of carbon is the energy released (-394 kJ) when 1 mole of graphite reacts with oxygen in the reaction,


C(graphite) + O2 = CO2,       dHo = -394 kJ. Since the measurement is done at the standard condition, a superscript o is usually placed on the right side of H in most literature. Incidentally, the above equation is for the formation of CO2, and the enthalpy of reaction happens to be the enthalpy of formation of CO2, designated as dHof = -394 kJ, as reviewed in the next paragraph.


As another example, when 1.0 mole Zn reacts with sufficient amount of HCl solution (1.0 M), 150 kJ is released. Thus, we write standard energy of reaction for Zn as,


Zn + 2 HCl(aq) = H2(g) + ZnCl2(aq),       dHo = -150 kJ.


The standard Enthalpy of Formation


Combination of elements at their standard states resulting in one compound is called a formation reaction. When enthalpy of formation is measured at the standard condition, it is called the standard enthalpy of formation. The standard enthalpy of combustion of carbon mentioned earlier C(graphite) + O2 = CO2,       dHof = -394 kJ. is the formation of CO2 from elements at their standard states. Thus, dHof of -394 kJ is also the standard enthalpy of formation of CO2


.


Similarly, a few more examples are given below. The enthalpies can be both positive and negative values.


1/2 O2 = O,       dHof = 249.17 kJ       (also bond energy of O=O).
1/2 H2 = H,       dHof = 217.96 kJ       (also bond energy of H-H).
H2 + O2 = H2O2(aq),       dHof = -191.17 kJ
1/2 H2 + 1/2 Cl2= HCl(g),       dHof = -92.31 kJ.
1/2 H2 + 1/2 Br2= HBr(g),       dHof = -36.40 kJ.


Hess's Law


Hess's law is another interpretation of the principle of conservation of energy. Since the changes in energy are independent of path, they depend on the initial and final state of the system. Thus, if it takes several steps to reach the final state from the initial state, the changes in energy are additative. The law states:


The total enthalpy change in a reaction is the same whether the reaction occurs in one or several steps.


However, one should recognize that the enthalpy change is related to the amounts of reactants and products in the equation.


A simple application of Hess's law is to give the standard enthalpy of decomposition of CO2 from its standard enthalpy of formation,


 


CO2(g) = C(graphite) + O2(g), dHo = 394 kJ.


Note that we change the sign of dHo if the reaction is reversed.

6 years ago
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