Calcium (Ca) donates two electrons to become Ca2+, and oxygen (O) each accepts an electron to become O2-. The Ca2+ and O2- ions combine in a 1:1 ratio to form the electrically neutral ionic compound, CaO, known as calcium oxide.
Explanation:Let's consider the Lewis symbols for each element to show the reaction of calcium and oxygen forming an ionic compound. Calcium (Ca), with an atomic number of 20, has two electrons in its outermost shell. Oxygen (O), with an atomic number of 8, has six electrons in its outermost shell and needs two more to achieve a full octet.
During the reaction, calcium donates its two electrons, one each going to two separate oxygen atoms. This results in calcium becoming a Ca2+ ion, losing its two outermost electrons. Upon receiving an electron, each oxygen atom becomes an O2- ion. The formation of these ions can be illustrated with Lewis symbols and arrows:
[Ca] 0 → [Ca]2+ + 2[e-]
[O]0 + [e-] → [O]2-
The positive charge of the calcium ion and the negative charges of the oxygen ions attract each other, and as a result, they come together to form the ionic compound calcium oxide, CaO. Since the ratio between Ca2+ and O2- needs to be 1:1 to balance the charges, the final compound formula is simply CaO, showing the combination of one calcium ion with one oxygen ion.
A 12.0 L gas cylinder is filled with 8.00 moles of gas. The tank is stored at 35°C. What is the pressure in the tank?
A.)The equilibrium constant, Kp, for the following reaction is 0.110 at 298 K:
NH4HS(s) NH3(g) + H2S(g)
Calculate the equilibrium partial pressure of H2S when 0.371 moles of NH4HS(s) is introduced into a 1.00 L vessel at 298 K.
PH2S = ?? atm
B.) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K:
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.47 atm at 500 K .
PPCl5 = ?? atm
PPCl3 = ?? atm
PCl2 = ?? atm
Final answer:
To calculate the equilibrium partial pressure of H2S, use the equilibrium constant expression and rearrange to solve for PH2S.
Explanation:
To calculate the equilibrium partial pressure of H2S, we need to use the equilibrium constant, Kp. The equilibrium constant expression for the given reaction is:
Kp = [H2S] / [NH4HS]
Given that Kp = 0.110, we can establish the relationship:
Kp = (PH2S) / (PNH4HS)
Since the initial moles of NH4HS is 0.371 moles and the volume of the vessel is 1.00 L, we can calculate the initial partial pressure of NH4HS:
PNH4HS = (0.371 mol) / (1.00 L) = 0.371 atm
Now, we can rearrange the equilibrium constant expression and solve for PH2S:
PH2S = Kp * PNH4HS = (0.110)(0.371 atm) = 0.0408 atm
The equilibrium partial pressure of H₂S is 0.332 atm. The equilibrium partial pressures for PCl₅, PCl₃, and Cl₂ are 0.83 atm, 0.64 atm, and 0.64 atm, respectively.
A.) To determine the equilibrium partial pressure of H₂S for the reaction NH₄HS(s) ⇌ NH₃(g) + H₂S(g) with Kp = 0.110 at 298 K, follow these steps:
Recognize that solid NH₄HS does not affect the equilibrium expression since its activity is 1.Let the equilibrium partial pressures of NH₃ and H₂S be PNH₃ = PH₂S = P since they are produced in a 1:1 ratio.Use the equation for Kp: Kp = PNH₃ × PH₂S = P².Substitute the given Kp value: 0.110 = P².Solve for P: P = √0.110 = 0.332 atm.Therefore, the equilibrium partial pressure of H₂S is 0.332 atm.
B.) For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with Kp = 0.497 at 500 K:
Set up an ICE table. Let the initial pressure of PCl₅ be 1.47 atm, and the changes in pressure be -x for PCl₅ and +x for both PCl₃ and Cl₂.At equilibrium: PCl₅ = 1.47 - x, PCl₃ = x, and Cl₂ = x.The equilibrium expression is: Kp = (PCl₃ × PCl₂) / PPCl₅.Substitute the values: 0.497 = (x × x) / (1.47 - x).Solve the quadratic equation: 0.497 = x² / (1.47 - x).Rearrange to get 0.497(1.47 - x) = x² which gives 0.73 - 0.497x = x².Rearrange into standard quadratic form: x² + 0.497x - 0.73 = 0.Use the quadratic formula x = [-b ± √(b² - 4ac)] / 2a where a = 1, b = 0.497, and c = -0.73.Solve for x to get two possible solutions, but only the positive value is physically meaningful: x ≈ 0.64 atm.Calculate the equilibrium pressures: PPCl₃ = 0.64 atm, PCl₂ = 0.64 atm, and PPCl₅ = 1.47 - 0.64 = 0.83 atm.Therefore, the equilibrium partial pressures are PPCl₅ = 0.83 atm, PPCl₃ = 0.64 atm, and PCl₂ = 0.64 atm.