JEE PYQ: Thermodynamics Question 13
Question 13 - 2020 (07 Jan Shift 1)
For the reaction: $\text{A}(\ell) \to 2\text{B}(g)$ $\Delta U = 2.1$ kcal, $\Delta S = 20$ cal K$^{-1}$ at 300 K. Hence $\Delta G$ in kcal is ___.
Show Answer
Answer: $-2.7$
Solution
$\Delta U = 2.1$ kcal $= 2.1 \times 10^3$ cal. $\Delta n_g = 2$. $\Delta H = \Delta U + \Delta n_g RT = 2.1 \times 10^3 + 2 \times 2 \times 300 = 2100 + 1200 = 3300$ cal. $\Delta G = \Delta H - T\Delta S = 3300 - 300 \times 20 = 3300 - 6000 = -2700$ cal $= -2.7$ kcal.
BETA
