Exams › NEET › Chemistry

ΔS = ΔH/T = 1.435 × 10³/273 = 5.260 cal mol⁻K

1. ΔS = ΔH/T
2. ΔS = ΔH/T × 273
3. ΔS = ΔH/T × 1.435 × 10³
4. ΔS = ΔH/T × 5.260 cal mol⁻K

Correct answer: ΔS = ΔH/T

Solution

The formula for entropy change is ΔS = ΔH/T, which matches the given calculation. The other options incorrectly modify the formula.

Related NEET Chemistry questions

• An ideal gas is compressed in a closed container its U?
• Internal energy does not include
• Assertion The change in internal energy and change in heat enthalpy does not depend upon the path by which changes are brought in. Reason Both \( \Delta U \) and \( \Delta H \) are path independent as U and H are state functions.
• Liquid benzene burns in oxygen according to: \( 2 C_{6} H_{6}+15 O_{2} \rightarrow 12 C O_{2}+6 H_{2} O \) If density of liquid benzene is \( 0.88 \mathrm{g} / \mathrm{cc} \) what volume of \( \boldsymbol{O}_{2} \) at \( \mathrm{STP} \) is needed to complete the combustion of 39 cc of liquid benzene?
• When a bomb calorimeter is used to determine the heat of reaction, which property of the system under investigation is most likely to remain constant?
• When heat is given to gas in an isothermal change, the result will be

⚔️ Practice NEET Chemistry free + battle 1v1 →