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Methanol can be made from "synthesis gas", produced by the following reaction.

CH4 (g) + H2O (g) → 3H2 (g) + CO (g)
For this reaction, ΔH = +210 kJ mol-1 and ΔS = +216 J K−1 mol-1 .
Use these values to explain why this reaction is not spontaneous at 298 K. Calculate the temperature at which it becomes spontaneous.

User Wallis
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1 Answer

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Step-by-step explanation:

To determine whether a reaction is spontaneous or not at a given temperature, we can use the Gibbs free energy change (ΔG) equation:

ΔG = ΔH - TΔS

Where:

ΔG is the Gibbs free energy change

ΔH is the enthalpy change

T is the temperature in Kelvin

ΔS is the entropy change

If ΔG is negative, the reaction is spontaneous, while if ΔG is positive, the reaction is non-spontaneous.

Given that ΔH = +210 kJ mol^(-1) and ΔS = +216 J K^(-1) mol^(-1), we can calculate the Gibbs free energy change at 298 K (25°C) as follows:

ΔG = ΔH - TΔS

ΔG = (+210 kJ mol^(-1)) - (298 K)(+216 J K^(-1) mol^(-1))

(Note: Conversion from J to kJ is necessary)

Let's convert the units and perform the calculation:

ΔG = (+210 kJ mol^(-1)) - (0.298 kJ)(+216 kJ mol^(-1))

ΔG = +210 kJ mol^(-1) - 64.368 kJ mol^(-1)

ΔG = -45.368 kJ mol^(-1)

The calculated value for ΔG at 298 K is -45.368 kJ mol^(-1), indicating that the reaction is not spontaneous at this temperature since ΔG is negative.

To find the temperature at which the reaction becomes spontaneous, we can rearrange the equation:

ΔG = ΔH - TΔS

T = (ΔH / ΔS)

Let's calculate the temperature:

T = (+210 kJ mol^(-1)) / (+216 J K^(-1) mol^(-1))

(Note: Conversion from kJ to J is necessary)

T = (210 kJ mol^(-1)) / (216 × 10^3 J K^(-1) mol^(-1))

(Note: Conversion from kJ to J)

T ≈ 972 K

Therefore, the reaction becomes spontaneous at approximately 972 K (or 699°C) and above, where the temperature allows ΔG to become negative.

User Andre Malan
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