Pemanasan natrium bikarbonat akan menghasilkan CO_(2) menurut reaksi: 2NaHCO_(3)(s)⇌Na_(2)CO_(3)(s)+CO_(2)(g)+H_(2)O(g) Jika pada 125°C nilai K_(p) untuk reaksi tersebut adalah 0,25 , maka tekanan parsial (atm) gas-gas yang terbentuk dalam sistem adalah.... B. 0,50 E. 4,00 C. 1,00 D. 2,99 A. 0,25

<p> Here, we are supposed to implement the Chemical Equilibrium in the form of Partial Pressures.<br />Partial Pressures in a reaction where gases are products are guided by the value of Kp, i.e., the equilibrium constant expressed in terms of partial pressures of the reactants and products. In this case, the question has provided us the value of the Equilibrium constant (Kp) itself, and when only one mole of the gaseous reactant breaks down, each gaseous product should have 1:1 stoichiometric ratio, hence indicated by the equal pressure for each product. So, we have to find the partial pressure of a single mole of the gaseous products, CO2 and H2O. This will be simply equivalent to the Kp when only one mole of each gaseous product is generated. </p>