The equilibrium constant in terms of pressures for the reduction of tungsten(IV) oxide to tungsten at...

The equilibrium constant in terms of pressures for the reduction of tungsten(IV) oxide to tungsten at 25 °C is K_p = 3.82×10^-4, corresponding to the reaction

WO₂(s) + 2CO(g) = (double arrow) W(s) + 2CO₂(g)

If the total pressure of an equilibrium system at 25 °C is 2.67 atm, calculate the partial pressures of CO(g) and CO₂(g).

P_CO = atm

P_CO₂ = atm

Solutions

Expert Solution

Few things we have to remember in this type of problems. The total pressure given in the problem is the sum of the partial pressure of the gas components. So here also the sum of the partial pressure of CO₂ and CO is the total pressure. Another thing is that the partial pressure applying by solid components is measured as 1. So, we can reduce the equilibrium constant in terms of pressure (K_p) by eliminating these terms. The partial pressure of both the gases are calculated here. We can do a back calculation also to check the answer if it is right or wrong.

queen_honey_blossom answered 4 months ago

Next > < Previous

Related Solutions

Suggest how the thermodynamic equilibrium constant and the equilibrium constant expressed in terms of partial pressures...

Suggest how the thermodynamic equilibrium constant and the equilibrium constant expressed in terms of partial pressures may respond differently to changes in temperature and pressure.

The equilibrium constant in terms of pressures, Kp, for the reaction NH3(g) + HF(g) NH4F(s) at...

The equilibrium constant in terms of pressures, Kp, for the reaction NH3(g) + HF(g) NH4F(s) at 25 °C is 10.2. (a) If the partial pressure of ammonia is PNH3 = 0.553 atm (at equilibrium) and solid ammonium fluoride is present, what is the equilibrium partial pressure in atm of hydrogen fluoride at 25 °C? An excess of solid NH4F is added to a container filled with NH3 at 25 °C and a pressure of 1.09 atm. Calculate the pressures of...

Consider the equilibrium Na2O(s) + SO2(g) ⇌ Na2SO3(s). (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures.

Consider the equilibrium Na2O(s) + SO2(g) ⇌ Na2SO3(s).(a) Write the equilibrium-constant expression for this reaction in terms of partial pressures.(b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.

The equilibrium constant, K, of a reaction at a particular temperature is determined by the concentrations or pressures of the reactants and products at equilibrium.

The equilibrium constant, K, of a reaction at a particular temperature is determined by the concentrations or pressures of the reactants and products at equilibrium. For a gaseous reaction with the general form aA+bB⇌cC+dD the Kc and Kp expressions are given by Kc=[C]c[D]d/[A]a[B]b Kp=(PC)c(PD)d(PA)a(PB)b The subscript c or p indicates whether K is expressed in terms of concentrations or pressures. Equilibrium-constant expressions do not include a term for any pure solids or liquids that may be involved since their...

Use the standard reduction potentials to calculate the equilibrium constant for each of the following reactions:...

Use the standard reduction potentials to calculate the equilibrium constant for each of the following reactions: Note if deltaG° for each reaction would be greater than or less than zero. Co2+(aq) + Sn(s) = Co(s) + Sn2+(aq) Ag+(aq) + Cr2+(aq) = Ag(s) + Cr3+(aq)

Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant...

Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K. You may want to reference (Pages 868 - 871) Section 20.5 while completing this problem. Part B 3Ce4+(aq)+Bi(s)+H2O(l)→3Ce3+(aq)+BiO+(aq)+2H+(aq) Express the equilibrium constant to two significant digits. part C. N2H5+(aq)+4Fe(CN)63−(aq)→N2(g)+5H+(aq)+4Fe(CN)64−(aq) Express the equilibrium constant to two significant digits.

Using standard reduction potentials, calculate the standard emf, free energy, and equilibrium constant at 298K for...

Using standard reduction potentials, calculate the standard emf, free energy, and equilibrium constant at 298K for each of the following reactions: a. Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s) b. 2 NO3-(aq) + 8 H+(aq) 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq) c. Fe(s) + 2 Fe3+(aq) → 3 Fe2+(aq) Ive posted this twice already and the answers have been really inconsistant

The element titanium occurs on earth as titanium (IV) oxide and is found mostly in the...

The element titanium occurs on earth as titanium (IV) oxide and is found mostly in the mineral called rutile. Rutile is abundant in beach sands in Australia and South Africa. For titanium to be useful, rutile has to be processed to obtain the basic metal form of titanium. Although the process is costly, titanium is gaining popularity as a material for various equipment due to its strength and light weight. One such application is in sports equipment. 1) Explain the...

Calcualte the equilibrium constant for the reaction using tabulated data and write the equilibrium constant for...

Calcualte the equilibrium constant for the reaction using tabulated data and write the equilibrium constant for this reaction in terms of partial pressure 2H2(g)+O2(g) <--> 2H2O(g) at 298k, 1 bar

Use cell potential to calculate an equilibrium constant. Calculate the cell potential and the equilibrium constant...

Use cell potential to calculate an equilibrium constant. Calculate the cell potential and the equilibrium constant for the following reaction at 298 K: Cu2+(aq) + Hg(l) = Cu(s) + Hg2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: -------

Subjects