Question

Determine the equilibrium constant for the reaction \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) when the reaction occurs at \(100 \mathrm{kPa}\) and \(2000 \mathrm{K} .\) The natural logarithms of the equilibrium constant for the reaction \(\mathrm{C}+2 \mathrm{H}_{2} \rightleftharpoons \mathrm{CH}_{4}\) and \(\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) at \(2000 \mathrm{K}\) are 7.847 and 23.839, respectively.

Short answer

Question: Determine the equilibrium constant for the following reaction at 100 kPa and 2000 K: $$\mathrm{CH}_{4}+2\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$ Given the natural logarithms of the equilibrium constants for the reactions: $$\mathrm{C}+2\mathrm{H}_{2} \rightleftharpoons \mathrm{CH}_{4}$$ $$\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}$$ at 2000 K are 7.847 and 23.839, respectively.

Step-by-step solution

Calculate the equilibrium constants from the natural logarithms

First, we need to calculate the equilibrium constants using the natural logarithms given for both reactions. Let the equilibrium constant of the first reaction be \(K_{1}\), and the equilibrium constant of the second reaction be \(K_{2}\): $$K_{1} = e^{7.847}$$ $$K_{2} = e^{23.839}$$

Combine the simultaneous reactions

To obtain the equilibrium constant of the desired reaction, we need to combine the two reactions given: $$\mathrm{C}+2\mathrm{H}_{2} \rightleftharpoons \mathrm{CH}_{4}$$ $$\mathrm{CH}_{4}+\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{CO}_{2}$$ First, let's reverse the first reaction so that we'll have \(\mathrm{CH}_{4}\) on the right-hand side: $$\mathrm{CH}_{4} \rightleftharpoons \mathrm{C}+2\mathrm{H}_{2}$$ Next, write the second reaction twice because we have \(2\) moles of \(\mathrm{H}_{2}\mathrm{O}\) and \(2\) moles of \(\mathrm{O}_{2}\) in the balanced reaction we want: $$2(\mathrm{CH}_{4}+\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{CO}_{2})$$ Now, add the reversed first reaction and the multiplied second reaction to obtain the desired reaction: $$\cancel{\mathrm{CH}_{4}} \rightleftharpoons \cancel{\mathrm{C}}+2\mathrm{H}_{2}$$ $$2(\cancel{\mathrm{CH}_{4}}+\cancel{\mathrm{C}}+2\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{CO}_{2})$$ Resulting in: $$\mathrm{CH}_{4}+4\mathrm{O}_{2} \rightleftharpoons 2\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$

Combine the equilibrium constants

Since we reversed the first reaction and have two of the second reactions, we need to use the following equation: $$K_{\text{desired}} = \frac{1}{K_{1}} * K_{2}^2$$ Put the values of \(K_{1}\) and \(K_{2}\) into this equation to obtain the desired equilibrium constant: $$K_{\text{desired}} = \frac{1}{e^{7.847}} * (e^{23.839})^2$$

Calculate the equilibrium constant for the desired reaction

Finally, we calculate the equilibrium constant for the desired reaction: $$K_{\text{desired}} = \frac{1}{e^{7.847}} * (e^{23.839})^2$$ $$K_{\text{desired}} = e^{-7.847} * e^{47.678}$$ $$K_{\text{desired}} = e^{39.831}$$ Now we have the equilibrium constant for the desired reaction at \(2000 \mathrm{K}\).