Answer to Question #167185 in General Chemistry for Mohammad Umer

Question

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

(i)    DH_f^° (CH₃OH)                                                                                                           [1 mark]

(ii)  DH_c^° (CH₃OH)                                                                                                           [1 mark]

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

 

CH₃OH (l)

O₂ (g)

CO₂ (g)

H₂O (l)

DH_f^° / kJ mol^-1

-238

0

-394

-286

 

             Use this data to calculate a value for the enthalpy of combustion, DH_c^°, 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).

Answer 1

a.) i)  $CH_2+H_2O → CH_3OH$

ii)$CH_3OH(l) + O_2(g) → f CO_2(g) + 2H_2O(l)$

b.) (1) $C(s) + 2H_2(g) + \dfrac{1}{2}O_2(g) → CH_3OH(l)$ $Δ_fH$ = = -238 kJ

(2) $C(s) + O_2(g) → CO_2(g)$ $Δ_fH$ = -394 kJ

(3)$H_2 (g) + O(g) → H_2O(l)$ $Δ_fH$ = -286 kJ

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

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

= - $728$ kJ