For the following reaction at 300 K A_2(~g)+3 ~B_2(~g) arrow 2 AB_3(~g) the enthalpy change is +15 kJ then the internal energy change is:

Correct answer: A. 19988.4 ~J A_2(~g)+3 ~B_2(~g) arrow 2 AB_3(~g) Δ n_ (g) =n_ (P) -n_ (R) =2-3-1=-2 Δ H=Δ U+Δ n_g R T 15 × 1000=Δ U-2 × 8.314 × 300 Δ U=15000+600 × 8.314 =15000+6 × 831.4 =15000+4988.4 Δ U=19988.4 ~J

NEET 2024

Exam: NEET
Year: 2024
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2024 · Chemistry · Thermodynamics (C)

For the following reaction at 300 K A_2(~g)+3 ~B_2(~g) arrow 2...

For the following reaction at 300 K

\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})

the enthalpy change is +15 kJ then the internal energy change is:

Options

A $\quad 19988.4 \mathrm{~J}$
B200 J
C1999 J
D1.9988 kJ

Correct answer

A. $\quad 19988.4 \mathrm{~J}$

Step-by-step solution

\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})

\Delta n_{(\mathrm{g})}=\mathrm{n}_{(\mathrm{P})}-\mathrm{n}_{(\mathrm{R})}

=2-3-1=-2

\Delta H=\Delta U+\Delta n_g R T

15 \times 1000=\Delta U-2 \times 8.314 \times 300

\Delta U=15000+600 \times 8.314

=15000+6 \times 831.4

=15000+4988.4

\Delta U=19988.4 \mathrm{~J}

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