Chemical Thermodynamics Question 238
Question: The enthalpy of combustion of benzene from the following data will be (i) $ 6C(s)+3H_2(g)\to C_6H_6(l);,\Delta H=+45.9,kJ $ (ii) $ H_2(g)+\frac{1}{2}O_2(g)\to H_2O(l);,\Delta H=-285.9,kJ $ (iii) $ C(s)+O_2(g)\to CO_2(g);,\Delta H=-393.5,kJ $
Options:
A) + 3172.8 kJ
B) - 1549.2 kJ
C) - 3172.8 kJ
D) - 3264.6 kJ
Show Answer
Answer:
Correct Answer: D
Solution:
Multiplying eq. (ii) by (iii) and eq. (iii) by (vi), and then add $ 6,C+3H_2+\frac{15}{2}O_2\to $
$ 6,CO_2+6H_2O;\Delta H=3218.7 $ Subtract eq. (i) from the above equation and find the required result.
