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Following two equilibria are established...

Following two equilibria are established on mixing two gases `A_(2)` and `C`.
i. `3A_(2)(g) hArr A_(6)(g) " " K_(p)=1.6 atm^(-2)`
ii. `A_(2)(g)+C(g) hArr A_(2)C(g)`
If `A_(2)` and `C` mixed in `2:1` molar, ratio calculate the equilibrium partial pressure of `A_(2)`, C, `A_(2)C` and `K_(p)` for the reaction (ii). Given that the total pressure to be `1.4` atm and partial pressure of `A_(6)` to be `0.2` atm at equilibrium

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To solve the problem step-by-step, we will analyze the given equilibria and use the provided information to calculate the equilibrium partial pressures of the gases involved and the equilibrium constant \( K_p \) for the second reaction. ### Step 1: Write the reactions and equilibrium expressions We have two reactions: 1. \( 3A_2(g) \rightleftharpoons A_6(g) \) with \( K_p = 1.6 \, \text{atm}^{-2} \) 2. \( A_2(g) + C(g) \rightleftharpoons A_2C(g) \) ...
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