Answer:
The reaction is spontaneous when T> 0.98 Kelvin OR T> -272.17°C
Explanation:
Step 1: Data given
ΔH = 131.3 kJ/mol = 131300 J/mol
ΔS = 133.6 J/K*mol
T = 298K
Step 2: The balanced equation
C (s) + H2O (g) --> CO (g) + H2 (g)
Step 3: ΔG
For a reaction to be spontaneous, ΔG should be <0
When ΔG > 0 the reaction is spontaneous in the reverse direction.
ΔG = ΔH - TΔS
Since ΔG<0
ΔH - TΔS <0
Step 4: Calculate T where the reaction is spontaneous
ΔH - TΔS <0
131300 J/mol - T*133.6 J/K*mol <0
- T*133.6 J/K*mol < -131300 J/mol
-T <-131300 /133.6
-T< -982.8 Kelvin
T> 982.8 Kelvin OR T> 709.6°C
The reaction is spontaneous when T> 982.8 Kelvin OR T> 709.6°C
At 298 K this reaction C (s) + H2O (g) --> CO (g) + H2 (g) is not spontaneous
Answer:
25°C
Explanation:
Looking at the reaction given, the reaction is endothermic. The rate of forward reaction increases with increase in temperature. We can also see from the information provided that the thermodynamic data for the reaction was measured at 298K. We must convert this to °C as follows, 298-273= 25°C. This implies that the thermodynamic data was obtained at 25°C. Since the reaction is endothermic (∆H is positive), under standard conditions, the reaction is spontaneous above 25°C since increase in temperature favours the forward reaction.
