In an endothermic reaction, if there is a loss in entropy the reaction will

1. Analyze the question

The reaction is endothermic, so ΔH is positive (ΔH > 0). There is a loss in entropy, so ΔS is negative (ΔS < 0). We are asked whether the reaction will be spontaneous.

2. The concept: Gibbs free energy

Spontaneity is decided by the Gibbs free energy change:

ΔG = ΔH − TΔS

• If ΔG < 0, the reaction is spontaneous.
• If ΔG > 0, the reaction is not spontaneous.
• If ΔG = 0, the system is at equilibrium.

3. Solution

• Endothermic: ΔH = positive (+)
• Entropy loss: ΔS = negative (−)
• Temperature T (in kelvin) is always positive.

So:

ΔG = (+) − T(−) = (+) + (+) = positive

Both terms push ΔG to be positive at every temperature. The enthalpy term is unfavourable (energy must be absorbed), and the entropy term is also unfavourable (the system becomes more ordered). Nothing can make ΔG negative, so the reaction cannot occur spontaneously at any temperature.

4. Answer

The reaction will not be spontaneous, which is option C.

Why the others are wrong:

• A: The result is not indeterminate. The signs of ΔH and ΔS clearly fix the sign of ΔG.
• B: Spontaneity needs ΔG < 0, which is impossible here.
• D: Equilibrium needs ΔG = 0, which would require ΔH = TΔS, and that cannot hold when ΔH is positive and ΔS is negative.

5. Exam tip

Remember the four sign combinations:

• ΔH −, ΔS +: always spontaneous
• ΔH +, ΔS −: never spontaneous (this question)
• ΔH −, ΔS −: spontaneous at low temperature
• ΔH +, ΔS +: spontaneous at high temperature

Note that "endothermic and entropy loss" is the "double bad" case, so the reaction never goes on its own.