Question

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from 300.0 \(\mathrm{K}\) to 350.0 \(\mathrm{K}\) . Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Short answer

The standard change in enthalpy for this reaction is approximately \(9275.7 \text{ J/mol}\), calculated using the Van't Hoff equation and the given temperature and equilibrium constant values.

Step-by-step solution

Identify the formula to be used

We need to find the standard change in enthalpy, and one equation relating it to temperature and the equilibrium constant is the Van't Hoff equation: \[\frac{d\ln K}{dT} = \frac{\Delta H^{\circ}}{R \cdot T^2}\] where K is the equilibrium constant, T is the temperature in Kelvin, R is the ideal gas constant, and ΔH° is the standard enthalpy change. However, as we are given the initial and final temperatures and the factor by which the equilibrium constant increases, we can rearrange the Van't Hoff equation into the following form: \[\ln \frac{K_2}{K_1} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Here, \(K_1\) and \(K_2\) are the initial and final equilibrium constants, respectively, and \(T_1\) and \(T_2\) are the initial and final temperature values.

Plug in the given values

We know that \(K_2 = 6.67 K_1\), \(T_1 = 300.0\) K and \(T_2 = 350.0\) K. First, let's substitute the given values into the equation: \[\ln \frac{6.67 K_1}{K_1} = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{350.0} - \frac{1}{300.0}\right)\]

Simplify and solve for ΔH°

Now, let's simplify the equation and solve for ΔH°: \[\ln 6.67 = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{350.0} - \frac{1}{300.0}\right)\] \[\Delta H^{\circ} = -R\left(\frac{1}{350.0} - \frac{1}{300.0}\right)^{-1} \ln 6.67\] We can now plug in the value of R, the ideal gas constant, which is \(8.314 \text{ J/mol K}\): \[\Delta H^{\circ} = -8.314 \left(\frac{1}{350.0} - \frac{1}{300.0}\right)^{-1} \ln 6.67\]

Evaluate the expression for ΔH°

Finally, calculate the value of ΔH°: \[\Delta H^{\circ} = -8.314 \cdot \frac{1}{\left(\frac{1}{350.0} - \frac{1}{300.0}\right)} \cdot \ln 6.67\] \[\Delta H^{\circ} \approx 9275.7 \text{ J/mol} \] So, the standard change in enthalpy for this reaction is approximately \(9275.7 \text{ J/mol}\).

Key concepts

Van't Hoff equation

The Van't Hoff equation is a fundamental tool in thermodynamics that relates the change in the equilibrium constant (K) of a chemical reaction to the change in temperature (T) and the standard change in enthalpy (ΔH°). This equation is particularly useful in predicting how the equilibrium position of a reversible reaction changes when the temperature varies, which can be crucial in chemical engineering and research.

Written in its differential form, the equation is:

• \[\frac{d\ln K}{dT} = \frac{\Delta H^{\circ}}{R \cdot T^2}\]

Where:

• \(d\ln K\) is the change in the natural log of the equilibrium constant,
• \(dT\) is the change in temperature,
• \(R\) is the ideal gas constant,
• \(T\) is the absolute temperature in Kelvin.
• This equation can be integrated to relate two states (initial and final) of a system, allowing us to find the enthalpy change based on observed equilibrium constants and temperatures.

Enthalpy change

Enthalpy change (ΔH) is a vital concept when examining reactions, indicating the heat absorbed or released at constant pressure. It's especially crucial when discussing the temperature dependence of chemical equilibria. In the context of the Van't Hoff equation, the enthalpy change is key to understanding how reactions respond to temperature changes.

For endothermic reactions, where ΔH is positive, increasing temperature typically pushes the equilibrium towards products. Conversely, for exothermic reactions, where ΔH is negative, increasing temperature tends to shift the equilibrium towards reactants. By using the Van’t Hoff equation in its rearranged form, one can algebraically determine the enthalpy change:

• \[\Delta H^{\circ} = -R \left(\frac{1}{T_2} - \frac{1}{T_1}\right)^{-1} \ln \frac{K_2}{K_1}\]

In this form:

• \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\), respectively,
• \(R\) is the ideal gas constant.

Equilibrium constant

The equilibrium constant ( K ) is a fundamental concept in chemical reactions, describing the ratio of concentrations of products to reactants at equilibrium. It quantitatively captures the balance of a reversible reaction, and its value provides insights into the position of equilibrium.

• When K is high, products predominate at equilibrium,
• Conversely, a low K value indicates that reactants are favored.

In temperature-dependent studies like the one using the Van't Hoff equation, changes in K can help infer enthalpy changes and predict reaction behavior under varying thermal conditions. When studying the temperature effect, it is crucial to note that a change in K hints at a temperature-induced shift in equilibrium, allowing scientists to subsequently calculate thermodynamic properties like ΔH° .

Ideal gas constant

The ideal gas constant (R) is a fundamental parameter in physics and chemistry, serving as a bridge between kinetic and thermodynamic properties of gases. In the context of chemical reactions and the Van't Hoff equation, R is used as a constant to relate changes in enthalpy and the equilibrium constant across different temperatures.

• Its value is approximately \(8.314 \text{ J/mol K}\),
• It ensures that equations like Van't Hoff's are dimensionally consistent.

Any thermodynamic equation that involves energy changes in chemical reactions typically features R due to its role in defining energy per mole per Kelvin. Understanding R helps in appreciating how energy interactions are calculated and predicted in chemical systems. This constant thus becomes a crucial part of the toolkit for studying the thermodynamics of chemical reactions.