For the reaction :
A($l$) $\to$ 2B(g)
$\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}$ at 300 K
Hence $\Delta$G in kcal is :
Answer (integer)
-2
Solution
A($l$) $\to$ 2B(g)
<br><br>We know,
$\Delta$H = $\Delta$U + $\Delta$n<sub>g</sub>RT
<br><br>and $\Delta$G = $\Delta$H - T$\Delta$S
<br><br>$\therefore$ $\Delta$G = $\Delta$U + $\Delta$n<sub>g</sub>RT - T$\Delta$S
<br><br>= 2.1 + ${{2 \times 2 \times 300} \over {1000}}$ - ${{300 \times 20} \over {1000}}$
<br><br>= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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