Question

Given the following reaction at equilibrium, if Kc = 1.90 × 1019 at 25.0 °C, Kp = ________.H2 (g) + Br2 (g) 2 HBr (g)A) 5.26 × 10-20B) 1.56 × 104C) 6.44 × 105D) 1.90 × 1019E) none of the above

Answer 1

In the given equilibrium reaction for the formation of HBr from H2 and Br2, the change in moles of gas is zero, therefore, the equilibrium constant Kp is equal to Kc. Since Kc is given as 1.90 × 10^19, Kp will also be 1.90 × 10^19.

Step-by-step explanation:

The subject of this question is Chemistry, specifically the concept of chemical equilibrium and the relationship between the equilibrium constants Kc and Kp. The equilibrium expression relates the concentrations of the reactants and products in a reaction at equilibrium. The relationship between Kc (equilibrium constant in terms of concentration) and Kp (equilibrium constant in terms of partial pressure) for a gaseous reaction at a constant temperature is given by the equation Kp = Kc(RT)^Δn, where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).

In the given reaction H2 (g) + Br2 (g) → 2 HBr (g), the change in moles of gas is 0 because there are two moles of gas on each side of the reaction. As a result, Δn is 0, and hence Kp equals Kc. Therefore, if Kc = 1.90 × 1019 at 25.0 °C, then Kp will be the same value, 1.90 × 1019.