Question

For a given reaction, ΔH = -26.6 kJ/mol and ΔS = -77.0 J/K⋅mol. The reaction is spontaneous ________. Assume that ΔH and ΔS do not vary with temperature. For a given reaction, = -26.6 kJ/mol and = -77.0 J/Kmol. The reaction is spontaneous ________. Assume that and do not vary with temperature. at T < 298 K at T > 345 K at all temperatures at T > 298 K at T < 345 K

Answer 1

Answer:

The reaction is spontaneous at T < 345 K

Explanation:

Step 1: Data given

ΔH = -26.6 kJ/mol

ΔS = -77.0 J/K*mol

Step 2: When is a reaction spontaneous:

When ΔG is:

ΔG<0 the process is spontaneous and may proceed in the forward direction ( From left to right; from reactants to products).

ΔG >0the process is non-spontaneous  but may proceed spontaneously in the reverse direction.

ΔG =0 the process is at equilibrium, with no net change taking place over time.

Step 3: ΔG

ΔG = ΔH -TΔS

with ΔH = the enthalpy

with T = the temperature

with ΔS = the entropy

⇒ When ΔS > 0 and ΔH < 0, the process is always spontaneous.

⇒ When ΔS < 0 and ΔH > 0, the process is never spontaneous, but the reverse process is always spontaneous.

⇒ When ΔS > 0 and ΔH > 0, the process will be spontaneous at high temperatures and non-spontaneous at low temperatures.

⇒ When ΔS < 0 and ΔH < 0, the process will be spontaneous at low temperatures and non-spontaneous at high temperatures.

Step 4: In this situation:

ΔH = -26.6 kJ/(mol) = -26600 J/(mol)

ΔS = -77.0 J/K*(mol)

this means when ΔS < 0 and ΔH < 0, the process will be spontaneous at low temperatures and non-spontaneous at high temperatures.

ΔG = -26600J/(mol) -T*-77.0 J/K*(mol)

To be spontaneous ΔG < 0

T <345.45 Kelvin

The reaction is spontaneous at T < 345 K