Question

The formation of methanol is important to the processing of new fuels. At \(298 \mathrm{~K}, K_{\mathrm{p}}=2.25 \times 10^{4}\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l) $$ If \(\Delta H_{\mathrm{rxn}}^{\circ}=-128 \mathrm{~kJ} / \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH},\) calculate \(K_{\mathrm{p}}\) at \(0^{\circ} \mathrm{C}\)

Short answer

Kp at 0°C is approximately 1.1354x10^17.

Step-by-step solution

- Identify the given data

The data provided are:1. The equilibrium constant at 298 K, denoted as 𝑲𝒑: \(K_{p}(298 \text{ K}) = 2.25 \times 10^{4} \)2. The change in enthalpy of the reaction, denoted as ΔH°: \(ΔH°_{\text{rxn}} = -128 \text{ kJ/mol} \)3. The initial temperature: 298 K4. The final temperature: 0°C, which converts to 273 K

- Utilize the Van’t Hoff equation

The Van’t Hoff equation relates the change in the equilibrium constant with temperature: \[\frac{d (\text{ln} K_p)}{dT} = \frac{ΔH°_{\text{rxn}}}{R T^2} \]

- Integrate the Van’t Hoff equation

To find the equilibrium constant at a temperature different from the initial, integrate the Van't Hoff equation: \[ \frac{d (\text{ln} K_p)}{dT} = \frac{ΔH°_{\text{rxn}}}{R T^2} \]Integrating both sides with respect to T from T1 to T2 gives: \[ \text{ln} \frac{K_{p2}}{K_{p1}} = -\frac{ΔH°_{\text{rxn}}}{R} \bigg( \frac{1}{T_2} - \frac{1}{T_1} \bigg) \]

- Substitute known values

Substitute the known values:\[ K_{p1} = 2.25 \times 10^4 \ ΔH°_{\text{rxn}} = -128 \times 10^3 \text{ J/mol } \ T_1 = 298 \text{ K } \ T_2 = 273 \text{ K } \ R = 8.314 \text{ J/mol·K } \]This yields: \[ \text{ln} \frac{K_{p2}}{2.25 \times 10^4} = -\frac{-128 \times 10^3 \text{ J/mol}}{8.314 \text{ J/mol·K}} \bigg( \frac{1}{273} - \frac{1}{298} \bigg) \]

- Calculate the right side

Calculate the expression on the right side of the equation:\[ -\frac{-128 \times 10^3}{8.314} \bigg( \frac{1}{273} - \frac{1}{298} \bigg) \]This value calculates as:\[ \frac{128 \times 10^3}{8.314} \bigg( \frac{1}{273} - \frac{1}{298} \bigg) = 211.33 \ - \frac{1}{273} + \frac{1}{298} = 0.000134 \ 211.33 \times 0.000134 = 28.3 \]

- Solve for \(\text{ln} \)

\(\text{ln}\frac{K_{p2}}{2.25 \times 10^4} = 28.3\)To solve for \(\frac{K_{p2}}{2.25 \times 10^4}\), use exponentiation with base \(e\):\(\text{e}^{28.3}\) is a large number; hence \(K_{p2} = (2.25 \times 10^4) \text{e}^{28.3}\).

- Final calculation

Calculate \(\text{e}^{28.3}\) and multiply by \(2.25 \times 10^4\).\(\text{e}^{28.3} \approx 5.0463\times10^{12}\)Thus,\(K_{p2} \approx 2.25 \times 10^4 \times 5.0463\times10^{12} = 1.1354\times10^{17}\)

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that studies the relationships between heat, temperature, and energy. In chemistry, it helps understand how energy changes during chemical reactions.
The Van’t Hoff equation, which we’ll discuss later, is rooted in thermodynamics. It highlights how equilibrium constants change with temperature.
When a chemical reaction occurs, it either absorbs or releases energy. This energy change is an essential aspect of thermodynamics. Imagine you’re holding ice. The energy from your warm hand melts the ice, turning it into water. This process involves energy transfer, a key concept in thermodynamics.

• System: The part of the universe we’re studying. For our exercise, it’s the reaction forming methanol.
• Surroundings: Everything else outside the system.
• State Functions: Properties like enthalpy (ΔH) and entropy (S) that depend only on the state of the system, not how it got there.

Chemical Equilibrium

Chemical equilibrium occurs when the rate at which reactants transform into products equals the rate at which products revert to reactants. For our methanol formation reaction, equilibrium is reached when the concentrations of CO, H₂, and CH₃OH levels off.
At equilibrium, the concentration of each reactant and product remains constant. This doesn’t mean the reactions stop; they simply occur at the same rate in both directions.
An important aspect of equilibrium is the equilibrium constant (K). For gas-phase reactions, we use the equilibrium constant K_p, which relates to the partial pressures of gases involved.
In our exercise, the equilibrium constant K_p at 298 K is given as 2.25 × 10³.

• If K_p is much greater than 1, products dominate at equilibrium.
• If K_p is much less than 1, reactants dominate at equilibrium.

Using the Van’t Hoff equation, we can calculate how K_p changes with temperature. This equation is handy for understanding how equilibrium shifts with varying conditions.

Enthalpy Change

Enthalpy change (\textbf{ΔH}) is the heat absorbed or released during a reaction at constant pressure. It’s a crucial concept in thermodynamics and our van’t Hoff equation.
For our methanol formation reaction, the enthalpy change, ΔH_rxn, is -128 kJ/mol. This negative value tells us the reaction is exothermic, meaning it releases heat.
When using the Van’t Hoff equation, ΔH helps determine how the equilibrium constant (K_p) changes with temperature.
Here’s how ΔH relates to the Van’t Hoff equation: \[ \frac{d (\text{ln} K_{p})}{dT} = \frac{\Delta H^{\circ}_{\text{rxn}}}{R T^{2}} \]
In the integrated form, this becomes: \[ \text{ln} \frac{K_{p2}}{K_{p1}} = -\frac{\Delta H^{\circ}_{\text{rxn}}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)\]
By substituting the given and known values:

• ΔH_rxn as -128 kJ/mol
• R (gas constant) as 8.314 J/mol·K
• T₁ as 298 K and T₂ as 273 K

We calculate K_p at 0°C (273 K). Understanding ΔH is key to predicting how temperature changes affect equilibrium.