JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

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)

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

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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: $727$

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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: $727$

JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry