Chapter 18: Problem 41
Calculate \(K_{P}\) for the following reaction at \(25^{\circ} \mathrm{C}:\) $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=2.60 \mathrm{~kJ} / \mathrm{mol} $$
Short Answer
Expert verified
The value of \( K_{P} \) is approximately 0.322.
Step by step solution
01
Convert ΔG° to Joules
First, convert the given standard Gibbs free energy change \( \Delta G^{\circ} = 2.60 \mathrm{~kJ/mol} \) into Joules per mole, since the formula we plan to use requires the energy unit in Joules. Thus, we have \( \Delta G^{\circ} = 2.60 \times 10^{3} \mathrm{~J/mol} \).
02
Use the Relation Between ΔG° and KP
Use the relation \( \Delta G^{\circ} = -RT \ln K_{P} \), where \( R = 8.314 \mathrm{~J/(mol \, K)} \), and \( T = 25^{\circ} \mathrm{C} = 298 \mathrm{~K} \) (since the temperature must be in Kelvin). Substitute these values along with \( \Delta G^{\circ} \) to solve for \( K_{P} \).
03
Substitute Known Values into the Formula
Substitute the known values into the equation: \( 2.60 \times 10^{3} = - (8.314)(298) \ln K_{P} \).
04
Solve for ln(KP)
Rearrange the equation to solve for \( \ln K_{P} \): \( \ln K_{P} = \frac{-2.60 \times 10^{3}}{8.314 \times 298} \). Calculate this value.
05
Calculate KP
Use the value of \( \ln K_{P} \) to find \( K_{P} \): \( K_{P} = e^{\ln K_{P}} \). Compute this to get the final result for \( K_{P} \).
06
Final Calculation
After calculating, you should find that \( K_{P} \approx 0.322 \). This completes the calculation based on the given \( \Delta G^{\circ} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
In the study of chemical reactions, the equilibrium constant, represented as \( K_{P} \) for gases, is a fundamental concept. It determines the ratio of the concentrations of products to reactants at equilibrium.
This ratio measures the extent to which a chemical reaction proceeds. A high \( K_{P} \) value indicates products are favored, whereas a low \( K_{P} \) suggests reactants are predominant.
The expression for \( K_{P} \) depends on the balanced chemical equation for the reaction:
This ratio measures the extent to which a chemical reaction proceeds. A high \( K_{P} \) value indicates products are favored, whereas a low \( K_{P} \) suggests reactants are predominant.
The expression for \( K_{P} \) depends on the balanced chemical equation for the reaction:
- For the given reaction of \( \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g) \), \( K_{P} \) involves the partial pressures of \( \mathrm{HI} \), \( \mathrm{H}_{2} \), and \( \mathrm{I}_{2} \).
- It's crucial to note the equation is defined at a specific temperature, as \( K_{P} \) is temperature-dependent.
Thermodynamics
Thermodynamics provides the backdrop for understanding energy changes in reactions, helping us connect these concepts to practical calculations. A primary parameter here is Gibbs Free Energy (
Here:
- \( \Delta G^{\circ} \) indicates if a reaction is spontaneous under constant pressure and temperature. If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous.
Here:
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.
Reaction Kinetics
While thermodynamics can tell us if a reaction can occur, reaction kinetics explains how fast it occurs. Kinetics focuses on the rate at which reactants turn into products. This is separate from equilibrium since a reaction might be favorable (thermodynamically) but slow (kinetically).
Temperature, pressure, and catalysts can influence rates, but do not affect the equilibrium constant. It's crucial to remember that:
Temperature, pressure, and catalysts can influence rates, but do not affect the equilibrium constant. It's crucial to remember that:
- While \( \Delta G^{\circ} \) implies a favored direction, kinetics determines the time scale.
- Fast kinetics doesn't imply high \( K_{P} \); both concepts must work together to fully understand reactions.
Chemical Equilibrium
Chemical equilibrium represents a state in a reaction where the rates of forward and reverse reactions are equal. This means the concentrations of reactants and products have reached a steady state.
- At equilibrium, the Gibbs free energy is minimized, making the system very stable.
- The expression for equilibrium constant \( K_{P} \) only includes gaseous reactants and products.
- Increasing pressure or concentration of reactants typically shifts equilibrium towards products.
- Temperature changes can shift the equilibrium constant \( \left(K_{P}\right) \), influencing reaction spontaneity.