For the reaction $Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$,the equilibrium constant $K_p = 64 \ atm^2$. Calculate the total pressure at equilibrium.
(16 ATM) The equilibrium expression for the reaction is $K_p = P_{CO_2} \times P_{H_2O}$.
Since the stoichiometry of the gaseous products $CO_2$ and $H_2O$ is $1:1$,their partial pressures at equilibrium will be equal,i.e.,$P_{CO_2} = P_{H_2O} = P$.
Thus,$K_p = P \times P = P^2$.
Given $K_p = 64 \ atm^2$,we have $P^2 = 64$,which gives $P = 8 \ atm$.
The total pressure at equilibrium is $P_{total} = P_{CO_2} + P_{H_2O} = 8 \ atm + 8 \ atm = 16 \ atm$.
Since the stoichiometry of the gaseous products $CO_2$ and $H_2O$ is $1:1$,their partial pressures at equilibrium will be equal,i.e.,$P_{CO_2} = P_{H_2O} = P$.
Thus,$K_p = P \times P = P^2$.
Given $K_p = 64 \ atm^2$,we have $P^2 = 64$,which gives $P = 8 \ atm$.
The total pressure at equilibrium is $P_{total} = P_{CO_2} + P_{H_2O} = 8 \ atm + 8 \ atm = 16 \ atm$.
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