Q25 / 75 · JEE Main 2026 (02 April Shift 1) · Chemistry
At the transition temperature $T$, $A \rightleftharpoons B$ and $\Delta G^0 = 105 - 35\log T$ where $A$ and $B$ are two states of substance $X$. The transition temperature in $°$C when pressure is $1$ atm is _______. (Nearest integer)
Correct answer: 727
Solution
At the transition temperature and $1$ atm pressure, the two states $A$ and $B$ are in equilibrium.
For a phase transition of a pure substance at $1$ atm, both states are in their standard states, so the reaction quotient $Q = 1$.
Using the relation $\Delta G = \Delta G^0 + RT \ln Q$, we get $\Delta G = \Delta G^0$.
At equilibrium, $\Delta G = 0$, which implies $\Delta G^0 = 0$.
Given $\Delta G^0 = 105 - 35\log T$, we can equate it to zero:
$105 - 35\log T = 0$
$35\log T = 105$
$\log T = 3$
$T = 10^3 = 1000$ K
Converting the temperature to Celsius:
$T(^{\circ}\text{C}) = 1000 - 273 = 727^{\circ}\text{C}$
Answer: $727$
For a phase transition of a pure substance at $1$ atm, both states are in their standard states, so the reaction quotient $Q = 1$.
Using the relation $\Delta G = \Delta G^0 + RT \ln Q$, we get $\Delta G = \Delta G^0$.
At equilibrium, $\Delta G = 0$, which implies $\Delta G^0 = 0$.
Given $\Delta G^0 = 105 - 35\log T$, we can equate it to zero:
$105 - 35\log T = 0$
$35\log T = 105$
$\log T = 3$
$T = 10^3 = 1000$ K
Converting the temperature to Celsius:
$T(^{\circ}\text{C}) = 1000 - 273 = 727^{\circ}\text{C}$
Answer: $727$
Hence, the correct answer is 727.