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Consider the following reaction at 248 C and 1.00 atm: CH3Cl (g) + H2(g) --> CH4(g)...

Consider the following reaction at 248 C and 1.00 atm: CH3Cl (g) + H2(g) --> CH4(g) + HCl (g)For this reaction, the enthalpy change at 248 C is -83.3 kJ/mol. At constant pressure the molar heat capacities (Cp) for the compounds are as follows: CH3Cl (48.5 J/mol K) ; H2 (28.9 J/mol K) ; CH4 (41.3 J/mol K) ; HCl (29.1 J/mol K) Assuming that the Cp values are independent of temperature, calculate ∆H for this reaction at 25 C.

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Answer #1

Solution-
HA <------------&gt; H+ + A-
Initial: 0.5 0.0 0.0
equal: 0.43 0.07 0.07
=>Ka = (0.07)^2/(0.43)
= 0.011
∆Cp = (41.3+29.1-48.5-28.9) = -7 JK^-1mol^-1
Now Cp = (ΔH/ΔT)p
So the
∆H° = -83.3 kJ/mol -∆Cp*(248°C-25°C)
= -83.3 kJ/mol + 7*223 J/mol
= 81.7 kJ/mol

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