logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

Compute the heat of formation of liquid methanol in kilojoules per mole, using the given data: \[\] $ \Delta H_{vap}$ of liquid methanol $= -38\: kJ/mol, \Delta H_f^{\circ}\: of\: H(g), C(g), O(g)$ respectively are $218, 715\: and\: 247\: kJ/mol$. Also, $ \Delta H_{C-H}, \Delta H_{C-O}, \Delta H_{O-H}$ being $415, 356, 463 \: kJ/mol$ respectively.

$\begin {array} {1 1} (A)\;228\: kJ/mol & \quad (B)\;-228\: kJ/mol \\ (C)\;266 \: kJ/mol & \quad (D)\;-266 \: kJ/mol \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
$C(graphite) + 2H_2(g) + \large\frac{1}{2}$$ O_2(g) \rightarrow CH_3OH(g) \rightarrow CH_3OH(l)$
$ \Delta H = [ \Delta H_fC(g) + 2\Delta H_{H-H} + \large\frac{1}{2}$$ \Delta H_{O=O}] – [\Delta H_{C-O} + 3\Delta H_{C-H} + \Delta H_{O-H}] - \Delta H_{cond}(CH_3OH)$
$ = [\Delta H_fC(g) + 4\Delta H_f(H) + \Delta H_f(O)] – [\Delta H_{C-O} 3\Delta H_{C-H} + \Delta H_{O-H}] - \Delta H_{cond}(CH_3OH)$
$= (715 + 4 \times 18 + 249) – (356 + 3 \times 415 + 463) – 38 = -266\: kJ/mol$
Ans : (D)
answered Mar 19, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...