Answer to Question #167185 in General Chemistry for Mohammad Umer

Question #167185

4.   (a)  Write equations to represent the following enthalpy changes.

(i)    DHf° (CH3OH)                                                                                                           [1 mark]

(ii)  DHc° (CH3OH)                                                                                                           [1 mark]

(b)   The following table gives some standard enthalpies of formation.

 

CH3OH (l)

O2 (g)

CO2 (g)

H2O (l)

DHf° / kJ mol-1

-238

0

-394

-286

 

             Use this data to calculate a value for the enthalpy of combustion, DHc°, of methanol, CH3OH.

                                                                                                                                                     [4 marks]

(c)  Would expect the value obtained in part (b) to differ if gaseous rather than liquid water, is formed. Explain your answer.                                                                                                     [2 marks]

(d)  In an experiment 0.92 g of methanol was burned and the heat given off used to raise the temperature of 250 g of water. The temperature rise was 12 °C. The specific heat capacity of water is 4.2 J K-1 g-1.

             Calculate a value for the enthalpy change of combustion for one mole of methanol. [4 marks]

           (e)       Suggest two reasons why the experimental value of the enthalpy of combustion obtained in part (d) is less negative than the value obtained in part (b).


1
Expert's answer
2021-03-01T04:56:11-0500

a.) i)  "CH_2+H_2O \u2192 CH_3OH"

ii)"CH_3OH(l) + O_2(g) \u2192 f CO_2(g) + 2H_2O(l)"


b.) (1) "C(s) + 2H_2(g) + \\dfrac{1}{2}O_2(g) \u2192 CH_3OH(l)" "\u0394_fH" = = -238 kJ


(2) "C(s) + O_2(g) \u2192 CO_2(g)" "\u0394_fH" = -394 kJ


(3)"H_2 (g) + O(g) \u2192 H_2O(l)" "\u0394_fH" = -286 kJ


Reverse equation (1) and multiply 3 by (2) and add all three


We get, "\\Delta_fH" = "238 - 394 - 572"

= - "728" kJ


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS