Thermodynamics Question 62
Question: The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of $ H _{2} $ at 1.5 atm pressure is $ \Delta H=-0.31kJ $ . The value of $ \Delta E $ will be
[DPMT 2004]
Options:
A) ?0.3024 kJ
B) 0.3024 kJ
C) 2.567 kJ
D) ?0.0076 kJ
Show Answer
Answer:
Correct Answer: A
Solution:
$ C _{2}H _{2}(g)+H _{2}(g)\to C _{2}H _{4}(g) $
$ \Delta ng=1-2=-1 $ ; $ \Delta H=-0.31KJmo{{l}^{-1}} $
$ P=1.5atm $ , $ \Delta V=-50mL=-0.050L $
$ \Delta H=\Delta E+P\Delta V $
$ -0.31=\Delta E-0.0076 $ ; $ \Delta E=-0.3024KJ $
