Q.75. At the transition temperature T, A B and where A and B are two states of substance X. The tr

Q: What is the correct answer for this exam Chemistry Q75?

The correct answer is 727. Sol. At the transition temperature, the two states (A and B) are in equilibrium. For a process at equilibrium under standard conditions, the standard Gibbs free energy change ($\Delta G°$) is zero. Given: $\Delta G° = 105 - 35 \log T$ Set $\Delta G° = 0$ at the transition temperature: $0 = 105 - 35 \log T$ $35 \log T = 105$ $\log T = \frac{105}{35}$ $\log T = 3$ To find T, we take the antilog:

Q: Q.75. At the transition temperature T, A B and where A and B are two states of substance X. The transition temperature in °C when pressure is 1 atm is _________ (Nearest integer)

The correct answer is 727. Sol. At the transition temperature, the two states (A and B) are in equilibrium. For a process at equilibrium under standard conditions, the standard Gibbs free energy change ($\Delta G°$) is zero. Given: $\Delta G° = 105 - 35 \log T$ Set $\Delta G° = 0$ at the transition temperature: $0 = 105 - 35 \log T$ $35 \log T = 105$ $\log T = \frac{105}{35}$ $\log T = 3$ To find T, we take the antilog:

Question

Q.75. At the transition temperature T, A  B and  where A and B are two states of substance X. The transition temperature in °C when pressure is 1 atm is _________ (Nearest integer)

FirstInTest · Accepted Answer

Sol. At the transition temperature, the two states (A and B) are in equilibrium. For a process at equilibrium under standard conditions, the standard Gibbs free energy change ($\Delta G°$) is zero.

Given: $\Delta G° = 105 - 35 \log T$

Set $\Delta G° = 0$ at the transition temperature:
$0 = 105 - 35 \log T$
$35 \log T = 105$
$\log T = \frac{105}{35}$
$\log T = 3$

To find T, we take the antilog:
$T = 10^3$ K
$T = 1000$ K

The question asks for the transition temperature in °C.
$T(°C) = T(K) - 273.15$
$T(°C) = 1000 - 273.15 = 726.85°C$

Rounding to the nearest integer, the temperature is 727°C.

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