Consider the following reaction at 248oC and 1.00 atm: CH3Cl(g) + H2(g) CH4(g) + HCl(g) For
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CH3Cl(g) + H2(g) → CH4(g) + HCl(g)
For this reaction, the enthalpy change at 248oC is 283.3 kJ/ mol. At constant pressure the molar heat capacities (Cp) for the compounds are as follows: CH3Cl (48.5 JK-1mol-1), H2 (28.9 JK-1 mol-1), CH4 (41.3 JK-1 mol-1), and HCl (29.1 JK-1 mol-1).
a. Assuming that the Cp values are independent of temperature, calculate ΔHo for this reaction at 25oC.
b. Calculate ΔHof for CH3Cl using data from Appendix 4 and the result from part a.
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a Using Hesss law and the equation H nC p T CH 3 Cl248C H 2 248C CH 4 248C HCl248C H 1 833 kJ CH 3 ...View the full answer
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