The Value Of $K_{\text{eq}}$ For The Equilibrium $H_2(g) + I_2(g) \longleftrightarrow 2HI(g)$ Is 784 At A Certain Temperature. What Is The Value Of \$K_{\text{eq}}$[/tex\] For The Equilibrium $\frac{1}{2} H_2(g)
Introduction
In chemistry, the equilibrium constant ($K_{\text{eq}}$) is a crucial concept that helps us understand the balance between reactants and products in a chemical reaction. It is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this article, we will explore the value of $K_{\text{eq}}$ for the equilibrium $H_2(g) + I_2(g) \longleftrightarrow 2HI(g)$ and how it relates to the equilibrium constant for the reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$.
What is $K_{\text{eq}}$?
K_{\text{eq}}$ is a dimensionless quantity that represents the equilibrium constant for a chemical reaction. It is calculated using the formula: $K_{\text{eq}} = \frac{[\text{products}]^{\text{stoichiometric coefficients}}}{[\text{reactants}]^{\text{stoichiometric coefficients}}}
where [products] and [reactants] are the concentrations of the products and reactants, respectively, and the stoichiometric coefficients are the numbers of moles of each substance in the balanced chemical equation.
The Equilibrium Constant for the Reaction $H_2(g) + I_2(g) \longleftrightarrow 2HI(g)$
The equilibrium constant for the reaction $H_2(g) + I_2(g) \longleftrightarrow 2HI(g)$ is given as 784 at a certain temperature. This means that at equilibrium, the concentration of HI is 784 times the concentration of H2 and I2.
The Equilibrium Constant for the Reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$
To find the equilibrium constant for the reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$, we need to use the relationship between the two reactions.
Relationship Between the Two Reactions
The two reactions are related by the following equation:
We can see that the first reaction is twice as many moles of HI as the second reaction. Therefore, the equilibrium constant for the second reaction is half the equilibrium constant for the first reaction.
Calculating the Equilibrium Constant for the Reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$
Using the relationship between the two reactions, we can calculate the equilibrium constant for the reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$ as follows:
Conclusion
In conclusion, the value of $K_{\text{eq}}$ for the equilibrium $H_2(g) + I_2(g) \longleftrightarrow 2HI(g)$ is 784 at a certain temperature. Using the relationship between the two reactions, we can calculate the equilibrium constant for the reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$ as 392.
References
- Atkins, P. W., & de Paula, J. (2010). Physical chemistry (9th ed.). Oxford University Press.
- Chang, R. (2010). Chemistry: The central science (11th ed.). McGraw-Hill.
- Levine, I. N. (2012). Physical chemistry (6th ed.). McGraw-Hill.
Further Reading
- For more information on equilibrium constants, see the following resources:
Related Topics
Tags
- Equilibrium constant
- Chemical reactions
- Stoichiometry
- Chemistry
- Equilibrium
Introduction
In our previous article, we explored the value of $K_{\text{eq}}$ for the equilibrium $H_2(g) + I_2(g) \longleftrightarrow 2HI(g)$ and how it relates to the equilibrium constant for the reaction $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \longleftrightarrow HI(g)$. In this article, we will answer some frequently asked questions about equilibrium constants in chemistry.
Q: What is the difference between $K_{\text{eq}}$ and $K_{\text{c}}$?
A: $K_{\text{eq}}$ and $K_{\text{c}}$ are both equilibrium constants, but they are calculated differently. $K_{\text{eq}}$ is calculated using the formula:
where [products] and [reactants] are the concentrations of the products and reactants, respectively, and the stoichiometric coefficients are the numbers of moles of each substance in the balanced chemical equation.
K_{\text{c}}$, on the other hand, is calculated using the formula: $K_{\text{c}} = \frac{[\text{products}]^{\text{stoichiometric coefficients}}}{[\text{reactants}]^{\text{stoichiometric coefficients}}} \times \frac{1}{[\text{solvent}]}
where [solvent] is the concentration of the solvent.
Q: What is the relationship between $K_{\text{eq}}$ and $K_{\text{c}}$?
A: The relationship between $K_{\text{eq}}$ and $K_{\text{c}}$ is given by the equation:
This means that $K_{\text{c}}$ is equal to $K_{\text{eq}}$ multiplied by the reciprocal of the concentration of the solvent.
Q: How do I calculate the equilibrium constant for a reaction?
A: To calculate the equilibrium constant for a reaction, you need to follow these steps:
- Write the balanced chemical equation for the reaction.
- Identify the products and reactants in the equation.
- Determine the stoichiometric coefficients for each substance in the equation.
- Calculate the concentrations of the products and reactants at equilibrium.
- Plug the values into the formula for $K_{\text{eq}}$ and solve for $K_{\text{eq}}$.
Q: What is the significance of the equilibrium constant?
A: The equilibrium constant is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a useful tool for predicting the direction of a reaction and the extent to which it will proceed.
Q: Can the equilibrium constant be used to predict the rate of a reaction?
A: No, the equilibrium constant cannot be used to predict the rate of a reaction. The rate of a reaction is determined by the rate constant, which is a separate quantity that is related to the equilibrium constant.
Q: How does temperature affect the equilibrium constant?
A: Temperature affects the equilibrium constant by changing the concentrations of the products and reactants. As temperature increases, the equilibrium constant also increases, which means that the reaction will proceed in the forward direction.
Q: Can the equilibrium constant be used to predict the stability of a reaction?
A: Yes, the equilibrium constant can be used to predict the stability of a reaction. If the equilibrium constant is large, it means that the reaction is stable and will not proceed in the reverse direction.
Q: What is the relationship between the equilibrium constant and the Gibbs free energy?
A: The relationship between the equilibrium constant and the Gibbs free energy is given by the equation:
where $\Delta G$ is the change in Gibbs free energy, $R$ is the gas constant, $T$ is the temperature, and $\ln K_{\text{eq}}$ is the natural logarithm of the equilibrium constant.
Conclusion
In conclusion, the equilibrium constant is a fundamental concept in chemistry that helps us understand the balance between reactants and products in a chemical reaction. By answering these frequently asked questions, we hope to have provided a better understanding of the equilibrium constant and its significance in chemistry.
References
- Atkins, P. W., & de Paula, J. (2010). Physical chemistry (9th ed.). Oxford University Press.
- Chang, R. (2010). Chemistry: The central science (11th ed.). McGraw-Hill.
- Levine, I. N. (2012). Physical chemistry (6th ed.). McGraw-Hill.
Further Reading
- For more information on equilibrium constants, see the following resources:
Related Topics
Tags
- Equilibrium constant
- Chemical reactions
- Stoichiometry
- Chemistry
- Equilibrium