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 <u>not spontaneous</u>
The equation for percent yield is

As such, the theoretical yield or expected yield is, in this case, 122.7g of Pb, the actual in this case is 107.9g so we have

Therefore, the percent yield of her reaction is 89.94%