3 questions! Assortment of Chem Problems! PLEASE HELP?! Please answer if youre 1

Question

3 questions! Assortment of Chem Problems! PLEASE HELP?! Please answer if youre 100% POSITIVE!

An equilibrium mixture contains 0.300 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

It is not .71!

Using the given data, determine the rate constant of this reaction.

Using the given data, calculate the rate constant of this reaction.

Explanation / Answer

1) let us calculate the Keq

Keq = Product of concentrations of products / Product of concentration of reactants

Keq = [CO2] [H2] / [CO] [H2O]

Keq = 0.3 X 0.3 / 0.2 X 0.2 = 2.25

The increase in concentration of [CO] = 0.3

assuming [H2] and [H2O] are same as previous

Keq = [CO2] 0.3 / 0.2 X 0.3 = 2.25

[CO2] = 0.675

So moles of Co2 needs to be added = 0.675 - 0.3 = 0.375

2) First we will calculate the order with respect to the given reactants

As per the given rate of reactions

order with respect to A = 2

Order with respect to B = 0

So Rate of reaction = K[A]^2

K = 0.0181 / [0.29]^2 = 0.215 mol^-1 L sec^-1

3) We will determine the order of reaction with respect to both the reactants

Order with respect to A = 1

Order with respect to B = 2

RAte = K [A] [B]^2

K = Rate / [A][B]^2

K = 0.0814 / 0.23 X(0.25)^2 = 5.66 mol^2 L^2 sec^-1

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