Question

In: Chemistry

For the reaction below, Kp = 29.95 at 800 K. Calculate the equilibrium partial pressures of...

For the reaction below, Kp = 29.95 at 800 K. Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are P_PCl5 = 0.3900 atm and P_PCl3 = 0.4300 atm. Assume Cl2 is 0 atm. PCL5 (g) > PCl3 (g) + Cl2 (g)

Expert Solution

Let's write the reaction once again:

PCl₅ ----------------------> PCl₃ + Cl₂

i) 0.39 0.43 0

eq) 0.39 - x 0.43+x x

Kp = pCl₂ pPCl₃ / pPCl₅

Now, let's solve for x:

29.95 = (0.43 + x) (x) / (0.39 - x)

29.95 (0.39 - x) = x (0.43 + x)

11.681 - 29.95x = x² + 0.43x --------> At this point, we can rearrange the equation to a 2nd grade equation:

x² + 0.43x + 29.95x - 11.681 = 0

x² + 30.38x - 11.681 = 0

x = -30.38 (30.38² + 4x1x11.681)^1/2 / 2x1

x = -30.38 31.14 / 2

x₁ = 0.38

x₂ = -34.76

The value we are going to take would be 0.38 so:

pCl5 = 0.39 - 0.38 = 0.01 atm

pCl₃ = 0.43 + 0.38 = 0.81 atm

pCl₂ = 0.38 atm

This wuld be the correct values, and it's just one way to solve this problem.

The other way is doing in a longer way by calculating Kc,applying this expression: Kc = Kp (RT)ⁿ then with the ideal gas equation PV = nRT or P = cRT calculate c to all the gases, and doing later the same procedure as before to calculate x, and then converting again into pressure. You can choose the way you like the most.

queen_honey_blossom answered 1 year ago

For the reaction below, Kp = 29.21 at 800K. Calculate the equilibrium partial pressures of the...

For the reaction below, Kp = 29.21 at 800K. Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are Ppcl5 = 0.488atm & Ppcl3 = .4900atm. PCL5(g) <----> PCL3(g) + Cl2(g) PCL5: ______ atm Cl2: _______ atm PCl3: ______ atm

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 ´ 10-7. 2...

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 ´ 10-7. 2 H2S(g) f 2 H2(g) + S2(g) A reaction vessel at 800 K initially contains 3.50 atm of H2S. If the reaction is allowed to equilibrate, what is the equilibrium pressure of H2?

At 850.0 K, the value of the equilibrium constant Kp for the hydrazine synthesis reaction below...

At 850.0 K, the value of the equilibrium constant Kp for the hydrazine synthesis reaction below is 0.2300. N2(g)+H2(g)---> N2H2(g) If a vessel contains an initial reaction mixture in which [N2] = 0.02000 M, [H2] = 0.03000 M, and [N2H2] = 1.000×10-4 M, what will the [N2H2] be when equilibrium is reached? ________M

At 850. K, the value of the equilibrium constant Kp for the hydrazine synthesis reaction below...

At 850. K, the value of the equilibrium constant Kp for the hydrazine synthesis reaction below is 0.2100. If a vessel contains an initial reaction mixture in which [N2] = 0.02000 M, [H2] = 0.03000 M, and [N2H2] = 1.000×10-4 M, what will the [N2H2] be when equilibrium is reached?

At 1177 K the partial pressures of an equilibrium mixture of H2O, H2 and O2 are...

At 1177 K the partial pressures of an equilibrium mixture of H2O, H2 and O2 are 0.0340, 0.0035 and 0.0013 atm, respectively. Given that the chemical equation for the equilibrium is, 2 H2O(g) <----> 2 H2(g) + O2(g) calculate Kp and Kc.

For the reaction below, determine the equilibrium partial pressure (in atm) of HBr at 252.70 K...

For the reaction below, determine the equilibrium partial pressure (in atm) of HBr at 252.70 K if the initial pressure of H2 is 2.83 atm. Assume that ΔH and ΔS do not vary with temperature. Report your answer to three significant figures. LiBr (s) + H2 (g) ⇌ LiH (s) + HBr (g) ΔHf° (kJ/mol) S° (J mol-1 K-1) LiBr -351.20 74.30 H2 0.00 130.68 LiH -90.50 20.00 HBr -36.29 198.70

Consider the reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kp=2.91×103 at 298 K In a reaction at equilibrium, the partial pressure...

Consider the reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kp=2.91×103 at 298 K In a reaction at equilibrium, the partial pressure of SO2 is 0.135 atm and that of Cl2 is 0.342 atm . Part A What is the partial pressure of SO2Cl2 in this mixture?

The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g)...

The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) + CO(g) Calculate Δ G o for the reaction. kJ/mol Calculate Δ G for the reaction when the partial pressures are PH2 = 0.22 atm, PCO2 = 0.72 atm, PH2O = 0.66 atm, and PCO = 1.16 atm.

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA...

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 5.65 atm, PB = 5.98 atm, PC = 5.27 atm, and PD = 4.83 atm. What is the standard change in Gibbs free energy of this reaction at 25 °C?

1.For the reaction, PCl5(g) <-----> PCl3(g) + Cl2(g) Kp = 24.6 at 500 K calculate the equilibrium...

1.For the reaction, PCl5(g) <-----> PCl3(g) + Cl2(g) Kp = 24.6 at 500 K calculate the equilibrium partial pressures of the reactants and products if the initial pressures are PPCl5 = 0.610 atm, PPCl3 = 0.400 atm and PCl2 = 0.000 atm. PPCl5 = PPCl3 = PCl2 = 2. H2O(g) + Cl2O(g) <-----> 2HClO(g) Kc = 0.14 at 298.15 K calculate the equilibrium concentrations of the reactants and products if the initial concentrations are [H2O(g)] = 0.00482 mol L-1, [Cl2O(g)] = 0.00482...