Question

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Please answer this question :-

From the data given below at 298 K for the reaction
CH4(g) + 2O2 (g) -------> CO2(g)+2H2O(l)

Calculate the enthalpy of formation of CH4 (g) at 298 K

{ Given : Δr H = -890.5 kJ mol^-1 , Δr H(H2O) = -286.0 kJ mol^-1 , ΔrH(CO2) = -393.5 kJ mol^-1 }

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Answer 1

Hello Anaida.....It's Your Friend here^_^

Question is  We have to find the e n t h a l p y of formation of C H 4

The equation for e n t h a l p y of formation of methane is :

C(s) + 2 H 2(g) -------> C H 4 (g) L e t this be Named as "EQ A"

Given e n t h a l p y of reaction of C O 2 = -393.5...L e t's write the eq for this....

C(s) + O 2(g) -------> C O 2(g)..... H 1 =-393.5..L e t this be EQ 1

Given the e n t h a l p y of reaction of H 20 is -286 , L e t's now write the eq for this.....

H 2(g) + 1/2 O 2(g) -------> H 2 0(g).... H 2 = -286...L e t this be EQ 2( Please do note that the values of e n t h a l p y of reaction given in the question is for 1 mole of product...So balance eq accordingly.) ..And That's why we don't balance the above eq as  H 2 + O 2 -------> 2 H 2 O...Because..There..We have 2 moles in the product side..So we balance it in the first way shown above.

Also given that...

CH 4(g) + 2 O 2 (g) -------> CO 2(g)+2 H 2 O(l)  H 3 =  -890.5..L e t this be EQ 3

So doing possible operations on EQ 1, 2 AND 3 ...We need to arrive at EQA

Comparing Eq 1 and Eq A...Both have 1 C atom on left side..So l e t's keep Eq 1 as it is. H 1 = - 393.5

Comparing Eq 2 and Eq A..Eq A has 2 Hydrogen..However...Eq 2 has only 1 Hydrogen..on left side...So Multiply Eq  2 with "Two." Delta H 2= 2* -286 = - 572

n d Now comparing Eq A And Eq 3...We see that In Eq A...C H 4 is to the right side ..And it is to the left side in Eq 3...So we reverse Eq 3...Which means..The sign reverses.... H 3 =  +890.5.

Now adding up all the Eq 1 , 2, and 3 after the operations we did above...Water, Oxygen and Carbon Dioxide on both sides would get cancelled out...

We get...e n t h a l p y of formation of C H 4 , Hf = -393.5 - 572 + 890.5 = -75

And So....the e n t h a l p y of formation of methane is -75  K J m o l-1