Kp for formation of ethane from hydrogen and ethylene is 5.5 × 10^18 atm^-1 and Kp for formation of — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write down the chemical equations for the given equilibrium processes We are given the following equilibrium reactions and their corresponding partial-pressure-based equilibrium constants (\(K_p\)) at a temperature of \(323\text{ K}\): 1. **Formation of ethane (\(\ce{C2H6}\)) from hydrogen (\(\ce{H2}\)) and ethylene (\(\ce{C2H4}\)):** \[\ce{H2(g) + C2H4(g) <=> C2H6(g)} \quad \text{with} \quad K_{p1} = 5.5 \times 10^{18}\text{ atm}^{-1}\] 2. **Formation of ethylene (\(\ce{C2H4}\)) from hydrogen (\(\ce{H2}\)) and acetylene (\(\ce{C2H2}\)):** \[\ce{H2(g) + C2H2(g) <=> C2H4(g)} \quad \text{with} \quad K_{p2} = 5 \times 10^{26}\text{ atm}^{-1}\] Step 2 - Define the target chemical equation The question asks for the equilibrium constant (\(K_{p3}\)) for the reaction between hydrogen and acetylene to form ethane. Acetylene (\(\ce{C2H2}\)) reacts with hydrogen (\(\ce{H2}\)) to undergo complete hydrogenation, yielding ethane (\(\ce{C2H6}\)): \[\ce{2H2(g) + C2H2(g) <=> C2H6(g)}\] Step 3 - Use Hess's Law of chemical equilibria to find the target constant By comparing the reactants and products of our given equations with the target equation, we can see that adding the first two reactions yields the target reaction: \[\begin{array}{rll} \ce{H2(g) + C2H4(g)} &\ce{<=> C2H6(g)} & \quad (K_{p1} = 5.5 \times 10^{18}\text{ atm}^{-1}) \\ \ce{H2(g) + C2H2(g)} &\ce{<=> C2H4(g)} & \quad (K_{p2} = 5 \times 10^{26}\text{ atm}^{-1}) \\ \hline \end{array}\] Canceling the intermediate ethylene (\(\ce{C2H4(g)}\)) from both sides of the added equation gives: \[\ce{2H2(g) + C2H2(g) <=> C2H6(g)}\] According to the rules of multiple equilibria, when two or more chemical equations are added together to produce a net equation, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual steps: \[K_{p3} = K_{p1} \times K_{p2}\] Step 4 - Substitute the given values and calculate the final result \[K_{p3} = \left(5.5 \times 10^{18}\text{ atm}^{-1}\right) \times \left(5 \times 10^{26}\text{ atm}^{-1}\right)\] \[K_{p3} = 27.5 \times 10^{44}\text{ atm}^{-2}\] \[K_{p3} = \boxed{2.75 \times 10^{45}\text{ atm}^{-2}}\] Step 5 - Explain each option * **(A) \(2.75 \times 10^{45}\text{ atm}^{-2}\):** Correct. This value is obtained by correctly identifying the target reaction as the sum of the two given intermediate reactions and multiplying their equilibrium constants. * **(B) \(1.1 \times 10^{-8}\):** Incorrect. This is the ratio of the two constants, \(\frac{K_{p1}}{K_{p2}} = \frac{5.5 \times 10^{18}}{5 \times 10^{26}} = 1.1 \times 10^{-8}\), which corresponds to a different chemical transformation. * **(C) \(9.09 \times 10^7\):** Incorrect. This is the inverse ratio of the constants, \(\frac{K_{p2}}{K_{p1}} = \frac{5 \times 10^{26}}{5.5 \times 10^{18}} \approx 9.09 \times 10^7\). * **(D) \(3.63 \times 10^{-46}\text{ atm}^2\):** Incorrect. This is the reciprocal of the correct product, which corresponds to the reverse target reaction: \(\ce{C2H6(g) <=> 2H2(g) + C2H2(g)}\).