Asked Nov 3, 2019

Calculate the reaction enthalpy, ΔH, for the following reaction:     CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the series of reactions that follow:

  1. C(s)+2H2(g)→CH4(g), ΔH =−74.8 kJ.
  2. C(s)+O2(g)→CO2(g), ΔH =−393.5 kJ.
  3. 2H2(g)+O2(g)→2H2O(g), ΔH =−484.0 kJ.
  4. H2O(l)→H2O(g), ΔH =44.0 kJ.

Expert Answer

Step 1

As per the question, given reactions are mentioned below:


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(1.) C(s)+2H, (g) CH, (g) (2.) C(s)+0,(g) CO, (g) (3.)2H, (g)+0,(g)->2H20(g) (4.)H,0()HO(g) AH74.8 kJ AH2=-393.5kJ AH,-484.0kJ AH4 = 44.0kJ

Step 2

The reaction for which enthalpy is to be calculated will get by first adding equation (2) and (3) then subtracting equation (1) and (4) from the added result, by first multiply equation (4) by 2.  On doing so, the result will be shown as follows:


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CH, (g)+20, (g)>CO, (g) +2H20() 4

Step 3

The above result is the reaction for which enthalpy is to be determined.



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Δacion Η ΔΗ, + ΔΗ, -ΔΗ-2ΔΗ, = -393.5 kJ +(-484.0 kJ)(-74.8 kJ)-2 (44.0 kJ) =-890.7kJ


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