The equilibrium partial pressures for a reaction are such that:

The partial pressures of the only two gaseous products are equal to each other.

The total partial pressure of the one gaseous reactant is .

The total pressure of the products is double the partial pressure of the reactant.

All coefficients for products are and the coefficient for the reactant is .

What is the value of ?

Booster class

General Chemistry 2

Skill

Solve problems involving equilibrium calculations

answer

watching

221

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

1 Sep 2020

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Join us on Discord

Chemistry Study Group

Join now

Related questions

Â± Relating Different Forms of the Equilibrium Constant

For chemical reactions where all reactants and products are in the gas phase the amount of each gas in the vessel can be expressed either as partial pressures or as concentrations. As such the equilibrium constant for a gas phase reaction can also be expressed in terms of concentrations or pressures. For the general reaction,

aA(g)+bB(g)âcC(g)+dD(g)

Kp=(PC)c(PD)d(PA)a(PB)b and Kc=[C]c[D]d[A]a[B]b.

It is possible to interconvert between Kp and Kc using

Kp=Kc(RT)În

where R=0.08314 L bar molâ1 Kâ1 and În is the difference in stoichiometric coefficients between gaseous products and gaseous reactants.

If we assume that the gaseous reactant and products are âidealâ and the partial pressures of the reactants and products were expressed in bar, then the thermodynamic equilibrium constant, K is the magnitude (or numerical value) of Kp.

Part A

For the reaction

2CH4(g)âC2H2(g)+3H2(g)

Kc=0.160 mol2 Lâ2 at 1621 âC. What is Kp for the reaction at this temperature?

Enter your answer numerically.

Part B

For the reaction

N2(g)+3H2(g)â2NH3(g)

Kp=3.80Ã10â3 barâ2 at 265 âC. What is Kc for the reaction at this temperature?

Enter your answer numerically.

At a certain temperature and a total pressure of 1.2 ATM, the partial pressures of an equilibrium mixture 2A (g) --><-- B (g)

Are P(A)=0.60atm and P(A)=0.60atm. If the constant K(p) is 1.67, and the total pressure increased to 1.5 ATM what would be the new partial pressures of A and B at equilibrium?