Use the following data to estimate a value of \(K_{\mathrm{p}}\) at \(1200 \mathrm{K}\) for the reaction \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) $$\begin{array}{l} \text { C(graphite) }+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \quad K_{\mathrm{c}}=0.64 \\ \quad \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \\ \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g}) \quad K_{\mathrm{c}}=1 \times 10^{8} \end{array}$$

Step by step solution

01

Identifying the Needed Reactions

Inspect given reactions and identify which ones can be manipulated algebraically (through reverse, multiply or addition/subtraction) to obtain the desired reaction \(2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g)\). The ones to be used are: \n \n C(graphite) + CO2(g) \rightleftharpoons 2CO(g) with \(K_c=0.64\); \n CO2(g) + H2(g) \rightleftharpoons CO(g) + H2O(g) with \(K_c=1.4\); \n C(graphite) + 0.5O2(g) \rightleftharpoons CO(g) with \(K_c=1 × 10^8\)

02

Manipulate the Reactions and the Corresponding Equilibrium Constants

In order to add the reactions to achieve the desired final reaction, reverse the first reaction:2CO(g) \rightleftharpoons C(graphite) + CO2(g) with \(K_c= 1/0.64\),keep the second reaction as is and multiply the third reaction by 2:2C(graphite) + O2(g) \rightleftharpoons 2CO(g) with \(K_c= (1 × 10^8)^2\). Since reversing a reaction inverts the equilibrium constant and changing the coefficient of the reaction elevates the \(K_c\) to that power.

03

Adding up Reactions and Corresponding Equilibrium Constants

Add up the reactions manipulated in step one:\[ 2CO(g) \rightleftharpoons C(graphite) + CO2(g) \]\[ CO2(g) + H2(g) \rightleftharpoons CO(g) + H2O(g) \]\[ 2C(graphite) + O2(g) \rightleftharpoons 2CO(g)\]This results in the desired final reaction: \[ 2H_2(g) + O2(g) \rightleftharpoons 2H_2O(g)\]In the same manner, add up the corresponding \(K_c\)s by multiplying them together, as when adding up reactions, the corresponding equilibrium constants multiply. Therefore, the final \(K_c\) for the desired reactions is: \( (1/0.64) * 1.4 * (1 × 10^8)^2 \).

04

Conversion from \(K_c\) to \(K_p\)

Since the question asks for the value of \(K_p\) and the resultant equilibrium constant achieved is \(K_c\), transform \(K_c\) to \(K_p\) using the formula \(K_p = K_c (RT)^{\Delta n}\) where \(R\) is the ideal gas constant (0.08314 L bar /mol K), \(T\) is the absolute temperature in Kelvin (1200 K in this case), and \(\Delta n\) is the change in stoichiometric number (sum of the stoichiometric coefficient of the gaseous products minus the sum of the stoichiometric coefficient of the gaseous reactants). Substituting the obtained \(K_c\) and given \(T\) value and with the \(\Delta n\) being 0 in this given reaction because the number of gaseous moles remains the same on both sides of the reaction, solve for \(K_p\).

05

Calculate \(K_p\)

As \(\Delta n\) is zero, the equation simplifies to \(K_p = K_c\). The final \(K_p\) will be \( (1//64) * 1.4 * (1 × 10^8)^2 \), after doing the math.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!