InterviewSolution
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InterviewSolution
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Determine the enthalpy change of the reaction. `C_(3)H_(5) (g) + H_(2) (g) rarr C_(2)H_(6) (g) + CH_(4) (g)` at `25^(@)C`, using the given heat of combustion value under standard conditions: Compound `H_(2) (g) CH_(4) (g) C_(2) H_(6) (g) C` (graphite) `Delta H^(@) (kJ//mol) -285.8 =890.0 - 1560.0 -393.5` The standard heat of formation of `C_(3)H_(8) (g)` is `-103.8 kJ//mol` |
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Answer» Correct Answer - `-55.7 kJ//mol` From the given data, we can get following equations. (i) `H_(2) + (1)/(2) O_(2) rarr H_(2) O, Delta H_(1) = - 285.8 kJ` (ii) `CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2) O, Delta H_(2) = -890 kJ` (iii) `C_(2)H_(6) + (7)/(2)O_(2) rarr 2CO_(2) + 3H_(2) O, Delta H_(3) = - 1560 kJ` (iv) `C (s) + O_(2) rarr CO_(2), Delta H_(4) = - 393.5 kJ` (v) `3C (s) + 4H_(2) rarr C_(3) H_(8) , Delta H_(4) = - 103.8 kJ` The required equation is `C_(3)H_(8) (g) + H_(2) (g) rarr C_(2) H_(6) (g) + CH_(4) (g) Delta H = ?` We can get the desired equation using the manipulations given below `[3 xx (iv) + 5 xx (i) ] - [(v) + (iii) + (ii)]` `Delta H = (3 Delta H_(4) + 5 Delta H_(1)) - (Delta H_(5) + Delta H_(3) + Delta H_(2))` `= [3 xx (-393.5) + 5 xx (-285.8)] - [-103.8 - 1560 - 590]` `= -55.7 kJ` mole |
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