Question:
The temperature in K at which \(\Delta G = 0\), for a given reaction with \(\Delta H = -20.5 \text{ kJ mol}^{-1}\) and \(\Delta S = -50.0 \text{ J K}^{-1} \text{ mol}^{-1}\) is:
The temperature in K at which \(\Delta G = 0\), for a given reaction with \(\Delta H = -20.5 \text{ kJ mol}^{-1}\) and \(\Delta S = -50.0 \text{ J K}^{-1} \text{ mol}^{-1}\) is:
Show Hint
Always double-check your unit scales! Mixing \(\text{kJ}\) and \(\text{J}\) is a classic trap. Notice that if you forgot to multiply by \(1000\), you would evaluate \(\frac{-20.5}{-50} = 0.41\), or if you inverted the fraction, you would get \(2.44\) (matching choice 3). Keeping track of your unit multipliers keeps these calculation slip-ups at bay.
Updated On: May 25, 2026
- \(-410\)
- \(410\)
- \(2.44\)
- \(-2.44\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
The spontaneity of a chemical process is governed by the Gibbs-Helmholtz Equation, which relates change in Gibbs free energy (\(\Delta G\)), change in enthalpy (\(\Delta H\)), and change in entropy (\(\Delta S\)) at a constant thermodynamic absolute temperature (\(T\)):
\[
\Delta G = \Delta H - T\Delta S
\]
When a system reaches dynamic equilibrium, the net change in Gibbs free energy is exactly zero (\(\Delta G = 0\)). Setting this condition allows us to solve for the specific threshold temperature:
\[
0 = \Delta H - T\Delta S \quad \Rightarrow \quad T = \frac{\Delta H}{\Delta S}
\]
Critical Requirement: The units for both thermodynamic energy variables must match before division (usually by transforming kilojoules into joules via \(1 \text{ kJ} = 1000 \text{ J}\)).
Step 1: Convert the given values to matching metric units.
We are given the following parameters:
• Enthalpy change (\(\Delta H\)) \(= -20.5 \text{ kJ mol}^{-1} = -20.5 \times 1000 \text{ J mol}^{-1} = -20500 \text{ J mol}^{-1}\)
• Entropy change (\(\Delta S\)) \(= -50.0 \text{ J K}^{-1} \text{ mol}^{-1}\)
Step 2: Substitute parameters into the equilibrium temperature formula.
Using the derived relation \(T = \frac{\Delta H}{\Delta S}\): \[ T = \frac{-20500 \text{ J mol}^{-1}}{-50.0 \text{ J K}^{-1} \text{ mol}^{-1}} \] Cancel out the negative signs in both the numerator and denominator: \[ T = \frac{20500}{50} \] \[ T = 410 \text{ K} \]
We are given the following parameters:
• Enthalpy change (\(\Delta H\)) \(= -20.5 \text{ kJ mol}^{-1} = -20.5 \times 1000 \text{ J mol}^{-1} = -20500 \text{ J mol}^{-1}\)
• Entropy change (\(\Delta S\)) \(= -50.0 \text{ J K}^{-1} \text{ mol}^{-1}\)
Step 2: Substitute parameters into the equilibrium temperature formula.
Using the derived relation \(T = \frac{\Delta H}{\Delta S}\): \[ T = \frac{-20500 \text{ J mol}^{-1}}{-50.0 \text{ J K}^{-1} \text{ mol}^{-1}} \] Cancel out the negative signs in both the numerator and denominator: \[ T = \frac{20500}{50} \] \[ T = 410 \text{ K} \]
Was this answer helpful?
0
0
Top BITSAT Chemistry Questions
- Which of the following silver salts is insoluble in water ?
- Which of the following is deliquescent ?
- Among the following sets of quantum numbers. Which one is incorrect for $4d$ electron ?
- In the purification of impure nickel by Mond's process, metal is purified by:
- When $25\, g$ of a non-volatile solute is dissolved in $100\,g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1}\, mm$. If the vapour pressure of water at $20^{\circ} C$ is $17.5\, mm$, what is the molecular weight of the solute?
View More Questions
Top BITSAT Work done in Thermodynamics Questions
- The work done by a gas during an isothermal expansion from 2 L to 4 L at a constant temperature, with pressure initially at 2 atm, is: (Use R = 8.314 J/mol·K, T = 300 K).
- What is the free energy change, Δ G, when 1.0 mole of water at 100C and 1 atm pressure is converted into steam at 100C and 1 atm pressure?
- For the reaction CO(g) + (1)/(2)textO₂(textg) → textCO₂(textg), which one of the statements is correct at constant T and P?
- For a given reaction, Δ H = 35.5kJ mol⁻1 and Δ S = 83.6J K⁻1mol⁻1. The reaction is spontaneous at (Assume that Δ H and Δ S do not vary with temperature)
- For a particular reversible reaction at temperature T, Δ H and Δ S were found to be both positive. If Tₑ is the temperature at equilibrium, then the reaction would be spontaneous when:
View More Questions
Top BITSAT Questions
- Two point charges $-q$ and $+ q$ are located at point's $(0, 0, - a)$ and, $(0, 0, a)$ respectively. The electric potential at a point $(0, 9, z)$, where $z > a$ is
- Which of the following must be known in order to determine the power output of an automobile?
- A large drop of oil (density $0.8 \,g / cm ^{3}$ and viscosity $\eta_{0}$ ) floats up through a column of another liquid (density $1.2\, g / cm ^{3}$ and viscosity $\eta_{L}$ ). Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on :
- A body is projected vertically upwards at time $ t = 0$ and it is seen at a height $H$ at time $t_1$ and $t_2$ second during its flight. The maximum height attained is ($g$ is acceleration due to gravity)
- At what point of a projectile motion, acceleration and velocity are perpendicular to each other ?
View More Questions