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Topic: Bond enthalpy and internal energy  (Read 8946 times)

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Offline jjkwest1

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Bond enthalpy and internal energy
« on: October 18, 2009, 10:23:13 PM »
1. Calculate the average H-S bond enthalpy in H2S(g) given the standard enthalpies of formation for H2S(g), H(g), S(g) as -20.1 kJ, 218kJ and 223kJ/mol respectively.
a. 461 kJ/mol
b. 679 kJ/mol
c. 340 kJ/mol
d. 10.1 kJ/mol
e. 231 kJ/mol

Alright, so i wrote the equation as

H2S --> 2H + S

and i found the enthalpy for the reaction as 679 kJ/mol...my questions is how do you get the average H-S bond? would you have to divide 679/2 since there is 2 moles of H and therefore get 340 kJ/mol?

2. The change in internal energy for combustion of 1 mole of CH4(g) in a cyclinder according to the reaction
                                            CH4(g) + 2O2(g) --> CO2(g) + 2H20(g) 
    is -836.3 kJ. If a piston connected to the cylinder performs 567 kJ of expansion work due to the combustion, how much heat is lost from the system( the reaction mixture)?

Would you use the equation
       delta E= q + w

-836.3= q + (-567)    q= -269 kJ

does that seem right? also since they ask for how much heat is lost can i just answer q= 269 kJ  instead of -269 kJ? thanks

Offline jjkwest1

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Re: Bond enthalpy and internal energy
« Reply #1 on: October 19, 2009, 12:53:19 PM »
1. Calculate the average H-S bond enthalpy in H2S(g) given the standard enthalpies of formation for H2S(g), H(g), S(g) as -20.1 kJ, 218kJ and 223kJ/mol respectively.
a. 461 kJ/mol
b. 679 kJ/mol
c. 340 kJ/mol
d. 10.1 kJ/mol
e. 231 kJ/mol

Alright, so i wrote the equation as

H2S --> 2H + S

and i found the enthalpy for the reaction as 679 kJ/mol...my questions is how do you get the average H-S bond? would you have to divide 679/2 since there is 2 moles of H and therefore get 340 kJ/mol?

2. The change in internal energy for combustion of 1 mole of CH4(g) in a cyclinder according to the reaction
                                            CH4(g) + 2O2(g) --> CO2(g) + 2H20(g) 
    is -836.3 kJ. If a piston connected to the cylinder performs 567 kJ of expansion work due to the combustion, how much heat is lost from the system( the reaction mixture)?

Would you use the equation
       delta E= q + w

-836.3= q + (-567)    q= -269 kJ

does that seem right? also since they ask for how much heat is lost can i just answer q= 269 kJ  instead of -269 kJ? thanks

Offline jjkwest1

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Re: Bond enthalpy and internal energy
« Reply #2 on: October 19, 2009, 07:38:43 PM »
Can somebody please help me... Thanks :)

Offline UG

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Re: Bond enthalpy and internal energy
« Reply #3 on: October 20, 2009, 11:01:21 PM »
1. Calculate the average H-S bond enthalpy in H2S(g) given the standard enthalpies of formation for H2S(g), H(g), S(g) as -20.1 kJ, 218kJ and 223kJ/mol respectively.
a. 461 kJ/mol
b. 679 kJ/mol
c. 340 kJ/mol
d. 10.1 kJ/mol
e. 231 kJ/mol

Alright, so i wrote the equation as

H2S --> 2H + S

and i found the enthalpy for the reaction as 679 kJ/mol...my questions is how do you get the average H-S bond? would you have to divide 679/2 since there is 2 moles of H and therefore get 340 kJ/mol?

How did you get this? You know that the  :delta:H = sum(formation of products) - sum(formation of reactants) ?


2. The change in internal energy for combustion of 1 mole of CH4(g) in a cyclinder according to the reaction
                                            CH4(g) + 2O2(g) --> CO2(g) + 2H20(g) 
    is -836.3 kJ. If a piston connected to the cylinder performs 567 kJ of expansion work due to the combustion, how much heat is lost from the system( the reaction mixture)?

Would you use the equation
       delta E= q + w

-836.3= q + (-567)    q= -269 kJ

does that seem right? also since they ask for how much heat is lost can i just answer q= 269 kJ  instead of -269 kJ? thanks
This just looks like a little maths to me. Looks correct, unless it is some kind a trick question :)  :-X

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