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 (ΔH°) for this reaction can be calculated using the van't Hoff equation: \[\frac{\Delta H^{\circ}}{R} = \frac{\ln{K_2} - \ln{K_1}}{{1}/{T_2}-{1}/{T_1}}\] Given the equilibrium constant ratio of 6.67 and temperatures 300.0 K and 350.0 K, we can substitute the values and solve for ΔH°: \[\Delta H^{\circ} ≈ -20807.7 \, J \cdot mol^{-1}\] Therefore, the standard change in enthalpy for this reaction is approximately -20.8 kJ/mol.

Step-by-step solution

Write the van't Hoff equation

The van't Hoff equation expresses the temperature dependence of the equilibrium constant (K), given the standard change in enthalpy: \[\frac{\Delta H^{\circ}}{R} = \frac{\ln{K_2} - \ln{K_1}}{{1}/{T_2}-{1}/{T_1}}\] Where, ΔH° is the standard change in enthalpy, K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively, and R is the gas constant(\(8.314 \, J \cdot mol^{-1} \cdot K^{-1}\)).

Find the ratio of the equilibrium constants

Given that the equilibrium constant increased by a factor of 6.67 when the temperature was increased from 300.0 K to 350.0 K, we can write: \[\frac{K_2}{K_1} = 6.67\]

Take the natural logarithm of the ratio of equilibrium constants

We need the difference between ln(K2) and ln(K1). We use the logarithmic property \(\ln{\frac{a}{b}} = \ln{a} - \ln{b}\) to obtain: \[\ln{\frac{K_2}{K_1}} = \ln{6.67}\]

Substitute values into the van't Hoff equation

We now have all the required values to substitute into the van't Hoff equation: \[\frac{\Delta H^{\circ}}{R} = \frac{\ln{6.67}}{(\frac{1}{350.0\,\mathrm{K}}) - (\frac{1}{300.0\,\mathrm{K}})}\]

Solve for ΔH°

To obtain the standard change in enthalpy, multiply both sides of the equation by R and solve for ΔH°: \[\Delta H^{\circ} = R \times \frac{\ln{6.67}}{(\frac{1}{350.0\,\mathrm{K}}) - (\frac{1}{300.0\,\mathrm{K}})}\] \[\Delta H^{\circ} = (8.314 \, J\cdot mol^{-1} \cdot K^{-1} ) \times \frac{\ln{6.67}}{(\frac{1}{350.0\,\mathrm{K}}) - (\frac{1}{300.0\,\mathrm{K}})}\] Evaluating this expression gives: \[\Delta H^{\circ} ≈ -20807.7 \, J \cdot mol^{-1}\] Thus, the standard change in enthalpy for this reaction is approximately -20.8 kJ/mol.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as K, is a crucial concept in chemistry that helps us understand how far a chemical reaction proceeds under a given set of conditions. It tells us the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients from the balanced equation, at equilibrium.
When we say a reaction has reached equilibrium, it means the rate of the forward reaction (products forming from reactants) is equal to the rate of the reverse reaction (reactants forming from products). This balance allows us to focus on concentrations rather than the rates of reaction.

• A large value of K (K > 1) indicates that at equilibrium, products are favored.
• A small value of K (K < 1) suggests that reactants are favored at equilibrium.
• If K equals 1, neither reactants nor products are favored at equilibrium.

The equilibrium constant is temperature-dependent, meaning that its value can change with temperature shifts.

Standard Change in Enthalpy

The standard change in enthalpy, denoted as \(\Delta H^{\circ}\), represents the heat change associated with a chemical reaction under standard conditions (1 atm pressure and concentrations of 1 mol/L for all reactants and products). It indicates whether a reaction is endothermic or exothermic.
For an endothermic reaction, \(\Delta H^{\circ}\) is positive, indicating that the reaction absorbs heat from the surroundings. Conversely, an exothermic reaction has a negative \(\Delta H^{\circ}\), signifying heat release.
This parameter is not only useful in understanding the energy profile of a reaction but also plays a key role in predicting how the equilibrium position shifts with temperature changes. The van't Hoff equation links \(\Delta H^{\circ}\) directly to how the equilibrium constant varies with temperature, giving insights into the thermodynamics of the reaction.

Temperature Dependence

Temperature profoundly influences both the rate of a reaction and the position of chemical equilibrium. According to the van't Hoff equation, the equilibrium constant changes when the temperature is altered.
The expression \(\ln{K_2} - \ln{K_1} = \frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\) encapsulates this temperature dependence, allowing us to calculate the change in the equilibrium constant as a function of temperature, given a constant \(\Delta H^{\circ}\).

• When you increase the temperature for an endothermic reaction, it tends to move towards the products. This shift increases K.
• For an exothermic reaction, raising the temperature leans the equilibrium towards the reactants, decreasing K.

Understanding these changes helps in predicting the direction of reaction shifts, a vital tool in industrial chemical processes.

Natural Logarithm

The natural logarithm, denoted by \(\ln\), is a mathematical function valuable in various scientific fields, including chemistry. It's specifically useful in the context of the van't Hoff equation. When dealing with equilibrium constants, the natural logarithm helps linearize data related to rate processes.
To simplify and analyze the relationship between temperature and the equilibrium constant, we use properties of the natural logarithm. For example, \(\ln(\frac{K_2}{K_1}) = \ln(K_2) - \ln(K_1)\), allows us to see how these ratios change with temperature shifts.
The nature of the \(\ln\) function aids in simplifying complex multiplicative relationships into additive ones, making computations far more straightforward. Any equations involving the \(\ln\) function can be more easily manipulated and solved, streamlining the process of handling data.