Question

In: Chemistry

It is known that at 300K the value of Kp for a certain reaction is 10^(12)....

It is known that at 300K the value of Kp for a certain reaction is 10^(12). For the reaction △H is 100kJ/mole.
a.Determine if this reaction is favorable at 800K and estimate Kp(800K). explain
b. The actual value of Kp (800) is 35. Explain any discrepancy from your estimate.

Expert Solution

Given and

(a) It is given that the reaction is feasible at 300K. That is according to the Van't Hoff equation , at 300K

so that . With increasing T the becomes much more negative. Hencea t higher temperature, e.g. 800K, the above inequality must holds and the free energy change will negative at this temperature. Therefore at 800K the reaction is feasible.

The temperature dependency of equilibrium constant is given by Van't Hoff isobar

where is the equilibrium constant at temperature and is at temperature . Now it is given that

and .

Therefore

This is the value of the equilibrium constant at 800K.

(b) The actual value of the equilibrium constant is .

Therefore it is observed that, there is a huge difference between the actual value and the value calculated above. This is due to the fact that, in Van't Hoff isobar the enthalpy change is taken to be independent of temperature. In reality this independency does not work well for a huge change in temperature e.g. from 300K to 800K. Hence at 800K must be different from at 300K.

queen_honey_blossom answered 2 years ago

a) Give the mathematical equilibrium constant expression Kc and Kp (not the value!) for the reaction...

a) Give the mathematical equilibrium constant expression Kc and Kp (not the value!) for the reaction 2N2O5(g) ↔ 4NO2 (g) + O2(g) b) At 250oC the concentrations of N2O5, NO2 and O2 are 0.100 mol/L, 0.0283 mol/L and 0.0105 mol/L. Calculate the values for Kc and Kp

17. What is the value of Kc for the following reaction given that Kp equals 6.48 x 10-6 (at 25 oC)?

Consider the following reaction and its corresponding equilibrium constant to answer the next two questions: P4 (g) + 6 H2 (g) Û 4 PH3 (g) Kp = 6.48 x 10-6 (at 25 oC) 17. What is the value of Kc for the following reaction given that Kp equals 6.48 x 10-6 (at 25 oC)? a) Kc = 3.98 x 10-12 b) Kc = 4.42 x 10-10 c) Kc = 5.60 x 10-5 d) Kc = 9.49 x 10-2 e) Kc...

1) At a certain temperature, 322 K, Kp for the reaction, Cl2(g) <=> 2 Cl(g), is...

1) At a certain temperature, 322 K, Kp for the reaction, Cl2(g) <=> 2 Cl(g), is 7.82 x 10-60. Calculate the value of ΔGo in kJ for the reaction at this temperature. 2) The equilibrium constant for the reaction, 2 Fe3+(aq) + Hg22+(aq) <=> 2 Fe2+(aq) + 2 Hg2+(aq) is 9.1 x 10-6(aq) at 298 K. Calculate ΔG in J when {Fe3+(aq)} = 0.375 {Hg22+(aq)} = 0.051 {Fe2+(aq)} = 0.021 {Hg2+(aq)} = 0.06

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 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?

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.140 at a certain temperature. A flask is charged...

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.140 at a certain temperature. A flask is charged with 0.500 atm PCl3 , 0.500 atm Cl2, and 0.300 atm PCl5 at this temperature. What are the equilibrium partial pressures of PCl3 , Cl2, and PCl5, respectively? Express your answers numerically in atmospheres with three digits after the decimal point, separated by commas.

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.160 at a certain temperature. A flask is charged...

For the exothermic reaction PCl3(g)+Cl2(g)⇌PCl5(g) Kp = 0.160 at a certain temperature. A flask is charged with 0.500 atm PCl3 , 0.500 atm Cl2, and 0.300 atm PCl5 at this temperature. What are the equilibrium partial pressures of PCl3, Cl2, and PCl5? How will the following changes affect the mole fraction of chlorine gas, X Cl2, in the equilibrium mixture. X Cl2 increases when __________ X Cl2 decreases when _________ X Cl2 stays the same __________ Fill in the blanks...

For the exothermic reaction PCl3(g)+Cl2(g)?PCl5(g) Kp = 0.200 at a certain temperature. A flask is charged...

For the exothermic reaction PCl3(g)+Cl2(g)?PCl5(g) Kp = 0.200 at a certain temperature. A flask is charged with 0.500 atm PCl3 , 0.500 atm Cl2, and 0.300 atm PCl5 at this temperature. What are the equilibrium partial pressures of PCl3 , Cl2, and PCl5, respectively? Express your answers numerically in atmospheres with three digits after the decimal point, separated by commas.

For the reaction Br2 (g) + Cl2 (g) → 2 BrCl Kp = 1.11 x 10-4...

For the reaction Br2 (g) + Cl2 (g) → 2 BrCl Kp = 1.11 x 10-4 at 150K. a) If 1.0 atm each of bromine and chlorine gas are placed in an otherwise empty 1 L container and allowed to equilibrate at a temperature of 150K, what will be the partial pressures at equilibrium of all three gases? Br2 atm Cl2 atm BrCl atm b) suppose 0.2 atm each of all three gases are placed in a selaed container and allowed to equilibrate...