For the reaction, \(2Cl(g)→Cl_2(g)\), what are the signs of \(∆H\) and \(∆S\) ?

Question

For the reaction
\(2A(g)+B(g)→2D(g)\)
\(∆U^Θ= -10.5\ kJ\) and \(∆S^Θ= -44.1\ JK^{-1}\).
Calculate \(∆G^Θ\) for the reaction, and predict whether the reaction may occur spontaneously.

Answer 1

Reaction

2 Cl(g) -> Cl2(g) Two chlorine atoms in the gas phase combine to form one chlorine molecule.

Sign of ΔH (Enthalpy Change)

In this process, a Cl–Cl bond is formed from two separate Cl atoms.
Bond formation releases energy (it is an exothermic process).

Therefore, ΔH is negative \((\Delta H < 0)\).

Sign of ΔS (Entropy Change)

Initially: 2 moles of gaseous particles (2 Cl(g)).
Finally: 1 mole of gaseous particles (Cl₂(g)).
The number of independent gas particles decreases, so the randomness/disorder of the system decreases.

Therefore, ΔS is negative \((\Delta S < 0)\).

Final Statement

For the reaction \(\ce{2 Cl(g) -> Cl2(g)}\):

ΔH < 0 (exothermic)
ΔS < 0 (entropy decreases)

For the reaction, \(2Cl(g)→Cl_2(g)\), what are the signs of \(∆H\) and \(∆S\) ?

Solution and Explanation

Reaction

Sign of ΔH (Enthalpy Change)

Sign of ΔS (Entropy Change)

Final Statement

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