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 $



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