Sodium carbonate can be made by heating sodium bicarbonate:
2NaHCO3(s) �¨ Na2CO3(s) + CO2(g) + H2O(g)
Given that ƒ¢H�‹ = 128.9 kJ/mol and ƒ¢G�‹ = 33.1 kJ/mol at 25�‹C, above what minimum temperature will the reaction become spontaneous under standard state conditions? Answer in celsius
33.1kj/mol =128.9kj/mol-T(dS) Put in T (273+25) and find dS.
then,
0=128.9Kj/mol-T ds use your found dS, then solve for T. Convert to Celcius
I got 216.52
To determine the minimum temperature at which the reaction becomes spontaneous under standard state conditions, we can use the relationship between enthalpy change (∆H) and Gibbs free energy change (∆G) given by the equation:
∆G = ∆H - T∆S
Where:
∆G = Gibbs free energy change
∆H = Enthalpy change
T = Temperature in Kelvin
∆S = Entropy change
We want to find the temperature at which ∆G is equal to zero (∆G = 0) to determine when the reaction becomes spontaneous. Rearranging the equation, we have:
0 = ∆H - T∆S
Solving for the temperature T:
T = ∆H / ∆S
Given that ∆H = 128.9 kJ/mol, we need to find the entropy change (∆S). Looking at the balanced chemical equation:
2NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(g)
We see that there are 3 moles of gaseous products (CO2 + H2O) and 1 mole of solid product (Na2CO3). The reaction reduces the number of moles from 2 (NaHCO3) to 1 (Na2CO3), resulting in a decrease in the entropy of the system. Hence, we can assume the entropy change (∆S) is negative (∆S < 0).
Now, we can substitute the values into the equation to find the minimum temperature:
T = 128.9 kJ/mol / ∆S
Given that ∆S = -33.1 kJ/mol, we substitute the value into the equation:
T = 128.9 kJ/mol / (-33.1 kJ/mol)
T = -3.9 K/mol
Since the standard temperature is 25°C (298 K), the minimum temperature at which the reaction becomes spontaneous under standard state conditions is:
Minimum temperature = 298 K - 3.9 K = 294.1 K
Converting to Celsius:
Minimum temperature in Celsius = 294.1 K - 273.15 = 20.95°C
Therefore, above a minimum temperature of approximately 20.95°C, the reaction will become spontaneous under standard state conditions.
To determine the minimum temperature at which the reaction becomes spontaneous under standard state conditions, we need to calculate the change in Gibbs free energy (ΔG) at different temperatures and find the temperature at which ΔG becomes negative.
The equation to calculate ΔG at a given temperature (T) is:
ΔG = ΔH - TΔS
Where:
ΔG = change in Gibbs free energy
ΔH = change in enthalpy
ΔS = change in entropy
T = temperature in Kelvin (K)
First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15:
25°C + 273.15 = 298.15 K
Given values:
ΔH = 128.9 kJ/mol
ΔG = 33.1 kJ/mol at 25°C (298.15 K)
Next, we need to calculate ΔS using the equation:
ΔS = ΔH / T
ΔS = 128.9 kJ/mol / 298.15 K
ΔS = 0.432 kJ/(mol·K)
Now, we can calculate ΔG at the given temperature:
ΔG = ΔH - TΔS
ΔG = 33.1 kJ/mol - 298.15 K * 0.432 kJ/(mol·K)
ΔG = 33.1 kJ/mol - 128.9288 kJ/mol
ΔG = -95.8288 kJ/mol
To find the minimum temperature at which the reaction becomes spontaneous (ΔG < 0), we can rearrange the equation:
ΔG = ΔH - TΔS
0 = ΔH - TΔS
TΔS = ΔH
T = ΔH / ΔS
Plugging in the values:
T = 128.9 kJ/mol / 0.432 kJ/(mol·K)
T = 298.65 K
Finally, we convert the temperature back to Celsius:
298.65 K - 273.15 = 25.5°C
Therefore, the reaction becomes spontaneous under standard state conditions above the minimum temperature of 25.5°C.