Using standard thermodynamic data from Appendix L of your text, calculate the equilibrium constant at 298 K for the following chemical reaction:
CO(g) + H2O(l) CO2(g) + H2(g).
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To calculate the equilibrium constant (K) for the given reaction, we will use the standard thermodynamic data.
The balanced equation for the reaction is:
CO(g) + H2O(l) → CO2(g) + H2(g)
The standard Gibbs free energy change (ΔG°) at 298 K is given by the formula:
ΔG° = ΣnΔG°(products) - ΣmΔG°(reactants)
First, we need to find the standard free energy change (ΔG°) of the reactants and products involved in the reaction.
From Appendix L of your textbook, we can find the standard free energy change values (ΔG°f) for each species.
The standard free energy change values at 298 K are:
ΔG°f(CO(g)) = -137.2 kJ/mol
ΔG°f(H2O(l)) = -237.2 kJ/mol
ΔG°f(CO2(g)) = -394.4 kJ/mol
ΔG°f(H2(g)) = 0 kJ/mol
Next, we substitute these values into the formula:
ΔG° = ΣnΔG°(products) - ΣmΔG°(reactants)
ΔG° = [ΔG°f(CO2(g)) + ΔG°f(H2(g))] - [ΔG°f(CO(g)) + ΔG°f(H2O(l))]
Now, let's substitute the values into the equation:
ΔG° = [-394.4 kJ/mol + 0 kJ/mol] - [-137.2 kJ/mol - 237.2 kJ/mol]
ΔG° = -394.4 kJ/mol + 137.2 kJ/mol + 237.2 kJ/mol
ΔG° = -20 kJ/mol
Now, we use the equation to calculate the equilibrium constant (K):
K = e^(-ΔG° / (RT))
where R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin (298 K).
Substituting the values into the equation:
K = e^(-(-20 kJ/mol) / (8.314 J/(mol·K) * 298 K))
K = e^(24020 J / (8.314 J/K*mol * 298 K))
K = e^(9.57)
Using the exponential function calculator or a calculator with an e^x function, the value of K comes out to be approximately 14675.7.
Therefore, the equilibrium constant (K) at 298 K for the given reaction CO(g) + H2O(l) → CO2(g) + H2(g) is approximately 14675.7.
To calculate the equilibrium constant (K) for the given reaction at 298 K, we need to use the standard thermodynamic data from Appendix L of your text.
The equilibrium constant (K) can be calculated using the equation:
K = exp(-ΔG°/RT)
Where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol∙K)), and T is the temperature in Kelvin.
1. First, we need to find the standard Gibbs free energy change (ΔG°) for the reaction. The standard Gibbs free energy change can be calculated using the equation:
ΔG° = ΣnΔG°f(products) - ΣmΔG°f(reactants)
Where ΔG°f is the standard Gibbs free energy of formation.
2. Look up the standard Gibbs free energy of formation values (ΔG°f) for each compound involved in the reaction from Appendix L of your text. The ΔG°f values are usually given in units of kJ/mol.
The ΔG°f values for the given compounds are:
ΔG°f(CO(g)) = -137.2 kJ/mol
ΔG°f(H2O(l)) = -237.2 kJ/mol
ΔG°f(CO2(g)) = -394.4 kJ/mol
ΔG°f(H2(g)) = 0 kJ/mol
3. Next, substitute the ΔG°f values into the equation for calculating ΔG°.
ΔG° = ΔG°f(CO2(g)) + ΔG°f(H2(g)) - ΔG°f(CO(g)) - ΔG°f(H2O(l))
ΔG° = (-394.4 kJ/mol) + (0 kJ/mol) - (-137.2 kJ/mol) - (-237.2 kJ/mol)
ΔG° = -157.2 kJ/mol
4. Convert the temperature to Kelvin. Given that the temperature is 298 K, no conversion is necessary.
5. Substitute the values into the equation for calculating K.
K = exp(-ΔG°/RT)
K = exp((-157.2 kJ/mol) / (8.314 J/(mol∙K) * 298 K))
K ≈ 11.16
Therefore, the equilibrium constant (K) at 298 K for the reaction CO(g) + H2O(l) → CO2(g) + H2(g) is approximately 11.16.