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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: malc7067 on November 20, 2006, 04:34:47 AM

Title: Energy Calculations
Post by: malc7067 on November 20, 2006, 04:34:47 AM

Hey again.

Ill just copy the question here, but basically I need someone to explain how this equation is solved using (products minus reactants) for formation enthalpies.

Calculate the average C - H bond energy for methane
DeltaHformation(kj/mol) CH4 = -75
DeltaHformation(kj/mol) C(gas) = 717
DeltaHformation(kj/mol) H(gas) = 218

this is where i keep getting stuck...

C(s) + 2H2(g) = CH4(g)

717 + 4(218) = X           Delta H= -75

When i find out X, by going X-(717+4(218))=-75, then divide by four, the answer is wrong... I was wondering if someone could do this problem and show me how you did it...
Also, i cant really understand how you can use the enthalpy of formation, then divide by 4, to find the bond energy????!?! I know there are 4 c-h in it, but just the fact that enthalpies of formation are used, seems strange...

Thanks.

Title: Re: Energy Calculations
Post by: DevaDevil on November 21, 2006, 10:31:58 PM

The full set of equations is:
C(s) + 2 H₂(g) --> CH4(g)   H^o = 1 mol x -75kJ/mol CH₄
C(g) --> C(s)  H^o = 1 mol x -717 kJ/mol C
 4 H(g)--> 2 H₂(g)   H^o = 4 mol x -218 kJ/mol H

Which make up the following reaction

C(g) + 4 H(g) --> CH4(g)    H^o = -1664 kJ

Hence, the bond breaking enthalpy of the C-H bond is 1664/4 = 416 kJ/mol