Chapter 14: Problem 57
The equilibrium reaction with methane as \(\mathrm{CH}_{4}\) \(\rightleftharpoons \mathrm{C}+2 \mathrm{H}_{2}\) has \(\ln K=-0.3362\) at \(800 \mathrm{~K}\) and In \(K=-4.607\) at \(600 \mathrm{~K}\). Noting the relation of \(K\) to temperature, show how you would interpolate In \(K\) in \((1 / T)\) to find \(K\) at \(700 \mathrm{~K}\) and compare that to a linear interpolation.
Short Answer
Step by step solution
Understand the Given and the Concept
Convert Temperatures to 1/T
Apply Linear Interpolation on ln K in Terms of 1/T
Direct Linear Interpolation on ln K
Comparison of Methods
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interpolation Methods
When dealing with temperature-dependent values like equilibrium constants, interpolation can be applied in two main ways:
- Using direct linear interpolation of the given property along a certain range.
- Using the reciprocal of the temperature ( 1/T ) as a basis for interpolation.
In the exercise, both methods were used to estimate ln K at 700 K. While in this case both methods gave the same result, using 1/T is generally seen as a reliable choice due to its alignment with non-linear thermodynamic principles.
Equilibrium Constant
The equilibrium constant can be expressed in various forms:
- For gaseous reactions, it often appears as K_p, referring to partial pressures.
- For those in solution, it takes the form of K_c, characterizing concentration.
Understanding ln K 's relationship with temperature allows chemists to predict how the equilibrium position shifts under varying thermal conditions, which is vital for reaction optimization in industrial and laboratory settings.
Temperature Dependence of K
If the reaction is exothermic, increasing temperature typically results in a decrease in K, indicating a shift towards the reactants. Conversely, for endothermic reactions, K increases with temperature, signifying a shift towards the products.
In the provided exercise, we have two known values of ln K at different temperatures. Using the interpolation methods demonstrated, students can see firsthand how to calculate ln K at an intermediate temperature (700 K) and understand that the equilibrium position could change when conditions shift. This information is essential for the practical application of thermodynamics in chemical processes.