SATHEE: Thermodynamics Question 62

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 $Δ H = - 0.31 k J$ . The value of $Δ E$ 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:

$C_{2} H_{2} (g) + H_{2} (g) \to C_{2} H_{4} (g)$

$Δ n g = 1 - 2 = - 1$ ; $Δ H = - 0.31 K J m o l^{- 1}$

$P = 1.5 a t m$ , $Δ V = - 50 m L = - 0.050 L$

$Δ H = Δ E + P Δ V$

$- 0.31 = Δ E - 0.0076$ ; $Δ E = - 0.3024 K J$

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Thermodynamics Question 62

Question: The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of H2 at 1.5 atm pressure is ΔH=−0.31kJ . The value of ΔE will be

Options:

Answer:

Solution:

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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 $Δ H = - 0.31 k J$ . The value of $Δ E$ will be