Calculate The Standard Free Energy Change In KJ For The Following Reaction At $25^{\circ} C$.$H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \quad K_{p} = 3.5 \times 10^{-1}$Your Answer:\$\square$[/tex\]

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Introduction

In chemistry, the standard free energy change (ΔG°) is a crucial concept that helps us understand the spontaneity of a reaction. It is a measure of the energy change that occurs when a reaction occurs under standard conditions. In this article, we will calculate the standard free energy change for the reaction: $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$ at $25^{\circ} C$, given the equilibrium constant (Kp) of 3.5 × 10^−1.

Understanding the Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a dimensionless quantity that can be used to predict the direction of a reaction and the extent to which it will occur. In this case, the equilibrium constant (Kp) is given as 3.5 × 10^−1.

Calculating the Standard Free Energy Change (ΔG°)

The standard free energy change (ΔG°) can be calculated using the following equation:

ΔG° = -RT ln(Kp)

where:

  • ΔG° is the standard free energy change
  • R is the gas constant (8.314 J/mol·K)
  • T is the temperature in Kelvin (298 K at 25°C)
  • ln(Kp) is the natural logarithm of the equilibrium constant (Kp)

Step 1: Convert the Temperature to Kelvin

First, we need to convert the temperature from Celsius to Kelvin. The temperature in Kelvin is given by:

T (K) = T (°C) + 273.15

T (K) = 25°C + 273.15 T (K) = 298.15 K

Step 2: Calculate the Natural Logarithm of the Equilibrium Constant (Kp)

Next, we need to calculate the natural logarithm of the equilibrium constant (Kp). The natural logarithm of 3.5 × 10^−1 is:

ln(Kp) = ln(3.5 × 10^−1) ln(Kp) = -2.25

Step 3: Calculate the Standard Free Energy Change (ΔG°)

Now, we can calculate the standard free energy change (ΔG°) using the equation:

ΔG° = -RT ln(Kp)

ΔG° = - (8.314 J/mol·K) × (298.15 K) × (-2.25) ΔG° = 5,813.5 J/mol

Step 4: Convert the Standard Free Energy Change (ΔG°) to kJ/mol

Finally, we need to convert the standard free energy change (ΔG°) from J/mol to kJ/mol. There are 1,000 J in 1 kJ, so:

ΔG° (kJ/mol) = ΔG° (J/mol) / 1,000 ΔG° (kJ/mol) = 5,813.5 J/mol / 1,000 ΔG° (kJ/mol) = 5.81 kJ/mol

Conclusion

In this article, we calculated the standard free energy change (ΔG°) for the reaction: $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$ at $25^{\circ} C$, given the equilibrium constant (Kp) of 3.5 × 10^−1. The standard free energy change (ΔG°) was calculated to be 5.81 kJ/mol.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Chang, R. (2010). Physical chemistry for the life sciences. Cambridge University Press.

Further Reading

  • For a more detailed explanation of the standard free energy change (ΔG°), see Atkins and De Paula (2010).
  • For a more detailed explanation of the equilibrium constant (Kp), see Chang (2010).

Introduction

In chemistry, the standard free energy change (ΔG°) is a crucial concept that helps us understand the spontaneity of a reaction. It is a measure of the energy change that occurs when a reaction occurs under standard conditions. In this article, we will calculate the standard free energy change for the reaction: $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$ at $25^{\circ} C$, given the equilibrium constant (Kp) of 3.5 × 10^−1.

Understanding the Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a dimensionless quantity that can be used to predict the direction of a reaction and the extent to which it will occur. In this case, the equilibrium constant (Kp) is given as 3.5 × 10^−1.

Calculating the Standard Free Energy Change (ΔG°)

The standard free energy change (ΔG°) can be calculated using the following equation:

ΔG° = -RT ln(Kp)

where:

  • ΔG° is the standard free energy change
  • R is the gas constant (8.314 J/mol·K)
  • T is the temperature in Kelvin (298 K at 25°C)
  • ln(Kp) is the natural logarithm of the equilibrium constant (Kp)

Step 1: Convert the Temperature to Kelvin

First, we need to convert the temperature from Celsius to Kelvin. The temperature in Kelvin is given by:

T (K) = T (°C) + 273.15

T (K) = 25°C + 273.15 T (K) = 298.15 K

Step 2: Calculate the Natural Logarithm of the Equilibrium Constant (Kp)

Next, we need to calculate the natural logarithm of the equilibrium constant (Kp). The natural logarithm of 3.5 × 10^−1 is:

ln(Kp) = ln(3.5 × 10^−1) ln(Kp) = -2.25

Step 3: Calculate the Standard Free Energy Change (ΔG°)

Now, we can calculate the standard free energy change (ΔG°) using the equation:

ΔG° = -RT ln(Kp)

ΔG° = - (8.314 J/mol·K) × (298.15 K) × (-2.25) ΔG° = 5,813.5 J/mol

Step 4: Convert the Standard Free Energy Change (ΔG°) to kJ/mol

Finally, we need to convert the standard free energy change (ΔG°) from J/mol to kJ/mol. There are 1,000 J in 1 kJ, so:

ΔG° (kJ/mol) = ΔG° (J/mol) / 1,000 ΔG° (kJ/mol) = 5,813.5 J/mol / 1,000 ΔG° (kJ/mol) = 5.81 kJ/mol

Conclusion

In this article, we calculated the standard free energy change (ΔG°) for the reaction: $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$ at $25^{\circ} C$, given the equilibrium constant (Kp) of 3.5 × 10^−1. The standard free energy change (ΔG°) was calculated to be 5.81 kJ/mol.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Chang, R. (2010). Physical chemistry for the life sciences. Cambridge University Press.

Further Reading

  • For a more detailed explanation of the standard free energy change (ΔG°), see Atkins and De Paula (2010).
  • For a more detailed explanation of the equilibrium constant (Kp), see Chang (2010).
    Calculating Standard Free Energy Change: A Q&A Guide ===========================================================

Introduction

In our previous article, we calculated the standard free energy change (ΔG°) for the reaction: $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$ at $25^{\circ} C$, given the equilibrium constant (Kp) of 3.5 × 10^−1. In this article, we will answer some frequently asked questions about calculating standard free energy change.

Q: What is the standard free energy change (ΔG°)?

A: The standard free energy change (ΔG°) is a measure of the energy change that occurs when a reaction occurs under standard conditions. It is a crucial concept in chemistry that helps us understand the spontaneity of a reaction.

Q: How is the standard free energy change (ΔG°) calculated?

A: The standard free energy change (ΔG°) can be calculated using the following equation:

ΔG° = -RT ln(Kp)

where:

  • ΔG° is the standard free energy change
  • R is the gas constant (8.314 J/mol·K)
  • T is the temperature in Kelvin (298 K at 25°C)
  • ln(Kp) is the natural logarithm of the equilibrium constant (Kp)

Q: What is the equilibrium constant (Kp)?

A: The equilibrium constant (Kp) is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a dimensionless quantity that can be used to predict the direction of a reaction and the extent to which it will occur.

Q: How do I convert the temperature from Celsius to Kelvin?

A: To convert the temperature from Celsius to Kelvin, you can use the following equation:

T (K) = T (°C) + 273.15

For example, if the temperature is 25°C, the temperature in Kelvin is:

T (K) = 25°C + 273.15 T (K) = 298.15 K

Q: How do I calculate the natural logarithm of the equilibrium constant (Kp)?

A: To calculate the natural logarithm of the equilibrium constant (Kp), you can use a calculator or a computer program. For example, if the equilibrium constant (Kp) is 3.5 × 10^−1, the natural logarithm is:

ln(Kp) = ln(3.5 × 10^−1) ln(Kp) = -2.25

Q: How do I convert the standard free energy change (ΔG°) from J/mol to kJ/mol?

A: To convert the standard free energy change (ΔG°) from J/mol to kJ/mol, you can divide the value by 1,000. For example, if the standard free energy change (ΔG°) is 5,813.5 J/mol, the value in kJ/mol is:

ΔG° (kJ/mol) = ΔG° (J/mol) / 1,000 ΔG° (kJ/mol) = 5,813.5 J/mol / 1,000 ΔG° (kJ/mol) = 5.81 kJ/mol

Q: What is the significance of the standard free energy change (ΔG°)?

A: The standard free energy change (ΔG°) is a measure of the energy change that occurs when a reaction occurs under standard conditions. It is a crucial concept in chemistry that helps us understand the spontaneity of a reaction. A negative value of ΔG° indicates that the reaction is spontaneous, while a positive value indicates that the reaction is non-spontaneous.

Conclusion

In this article, we answered some frequently asked questions about calculating standard free energy change. We hope that this article has been helpful in understanding the concept of standard free energy change and how to calculate it.

References

  • Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
  • Chang, R. (2010). Physical chemistry for the life sciences. Cambridge University Press.

Further Reading

  • For a more detailed explanation of the standard free energy change (ΔG°), see Atkins and De Paula (2010).
  • For a more detailed explanation of the equilibrium constant (Kp), see Chang (2010).