Question

For the reaction C (s) + H2O (g) --> CO (g) + H2 (g) ΔH = 131.3 kJ/mol and ΔS = 133.6 J/K-mol at 298K. At temperatures greater than _____ C this reaction is spontaneous under standard conditions.

Answer 1

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 2

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.