The value of enthalpy change $(Δ H)$ for the reaction $C_{2} H_{5} OH (l) + 3 O_{2} (g) ⟶ 2 {CO}_{2} (g) + 3 H_{2} O (l)$ at 27 °C is $1366.5 kJ {mol}^{- 1}$ . The value of internal energy change for the above reaction at this temperature will be

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$- 1369.0 kJmol - 1$

$- 1361.5 kJ {mol}^{- 1}$

$- 1371.5 kJ {mol}^{- 1}$

$- 1364.0 kJ {mol}^{- 1}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Since $Δ_{r} H = Δ_{r} U + (Δ v_{g}) R T,$ we have $Δ_{r} U = Δ_{r} H - (Δ v_{g}) R T$

For the given reaction $(Δ v_{g}) = - 1$ . Hence

$\begin{aligned} Δ_{r} U = - 1366.5 kJ {mol}^{- 1} - (- 1) (8.314 \times 10^{- 3} {kJK}^{- 1} {mol}^{- 1}) (300 K) \\ = (- 1366.5 + 2.494) kJ {mol}^{- 1} ≃ 1364.0 kJ {mol}^{- 1} \end{aligned}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!