Unit 4 questions

[lanvins]

Another question...

Carbon disulfide gas (CS2), which is used in the manufactre or rayon, can be made using an endothermic gas-phase reaction between sulfur trioxide gas(SO3) and carbon dioxide. Oxygen gas is also produced in the reaction

1. An equilibrium mixture of these gases was made by mixing sulfur trioxide and carbon trioxide. It consisted of .028mol of CS2, .022mol of SO3 and .014mol of CO2 in a 20L vessel. Calculate the value of the equilibrium constant at that temperature.

I don't get how to do it when they don't give you the mole of oxygen gas

[Mao]

1. An equilibrium mixture of these gases was made by mixing sulfur trioxide and carbon trioxide. It consisted of .028mol of CS2, .022mol of SO3 and .014mol of CO2 in a 20L vessel. Calculate the value of the equilibrium constant at that temperature.

hence, the amount of carbon disulfide and oxygen are in stoichiometrical ratios to each other, as only the reactants were present at the start.

[lanvins]

i think you got the balanced equation wrong, but i used ur method and got the right answer

[Mao]

oops [i've made so many mistakes tonight....]

[lanvins]

out of curiosity,

when you do stoichiometric ratios and you want to find an unkown mole, do you always have to use the mole of known product with unknown product and  known reactant with unknown reactant.

For example 2HI(g)------->H2(g) + 4I2(g) and you have the mole of HI and H2 and you want the mole of I2, why can't you multiply the mole of HI by 2 to find the unknown mole of I2.

Thanks, i hope that makes sense

[Mao]

out of curiosity,

when you do stoichiometric ratios and you want to find an unkown mole, do you always have to use the mole of known product with unknown product and  known reactant with unknown reactant.

For example 2HI(g)------->H2(g) + 4I2(g) and you have the mole of HI and H2 and you want the mole of I2, why can't you multiply the mole of HI by 2 to find the unknown mole of I2.

Thanks, i hope that makes sense

yes, you have to use something on the same side. this is only valid if you only had reactants to start off with and nothing else

that is, some reactants react to form products and the system comes to equilibirum. since the reaction produce products strictly to the mole ratio, hence all products formed will be at stoichiometric ratio.

the concentration of reactants at equilibrium is simply the "left over" unreacted stuff, which is not related to the amount of product present [at least not a simple relationship], hence it cannot be used.

all these rest on the assumption that the initial level of product is 0, that is, all products were formed by the forward reaction, hence they exist at stoichiometric ratio to each other. this is only valid for products formed in this way.

[bec]

Solutions of the acid–base indicator methyl red contain an equilibrium mixture of the red-coloured acid form of the indicator and a yellow-coloured conjugate base form. The equilibrium can be represented by:

HMr(aq) <--> H+(aq) + Mr–(aq)
  (red)      (yellow)

The equilibrium constant for the reaction is 10–5 M at 25°C.

What colour would the indicator be in solutions of:
i. pH 5?
ii. pH 10?

How do you work this out?

[Mao]

looking at K, its small values indicates the equilibrium is to the left.

at pH 5, the presence of H+ will drive the equilibrium further to the left than it already is, hence it'll be red.

at pH 10, the solution is basic, and will react with H+ to neutralise. hence, some H+ is removed from the RHS, and the equilibrium is driven to the right. now, assuming that we're using it as an indicator, it'll be present in very low concentrations, that is, we can treat this as an open system where H+ is continuously removed by the base, hence the equilibrium is driven to the right, and is yellow.

[bec]

thanks mao

[bec]

Does anyone have NEAP? Page 10, question 1.2.14, I don't understand why the number of moles of A₂ at equilibrium would be 0.1 (like it says in the answers), and not 0.2/3?

[lauzy358]

I have a question, mainly out of curiosity really.

For:
[Co(H2O)6]2+ + Cl- = [CoCl4]2- + H2O

Why does cobalt-water complez have a charge of 2+ and the cobalt-chloride complex have a charge of 2-?

I'm assuming it's 'cause H2O carries a positive charge, but I always figured water was neutral, so it must have to do with the bonding...? =S

ty in advance!

[Collin Li]

Cobalt has an oxidation state of +2, and exhibits a +2 charge.

With 6 neutral water ligands, the overall charge is +2.
With 4 negatively charged chloride ion ligands, the overall charge is (+2-4) = -2

[lauzy358]

Oh, thanks coblin ^^;
Didn't turn out to be as complicated as I thought =P

[Mao]

Does anyone have NEAP? Page 10, question 1.2.14, I don't understand why the number of moles of A₂ at equilibrium would be 0.1 (like it says in the answers), and not 0.2/3?

the starting amounts of the species are not in stoichiometrical ratios, that is, they do not have to be within stoichiometrical ratios

however, the following are true:

- the amount each chemical change by [for a particular reaction] is in stoichiometrical ratio to each other, as the species themselves react according to the stoichiometrical ratios as specified by the reaction formula

- consequently [as a particular case of the above], if we start with zero products, then the products produced will be in stoichiometrical ratios to other products [but not necessarily reactants]


so, for this particular reaction, B has changed by -0.6 mol, hence A₂ will change by -0.6/3 = -0.2 mol. as it started at 0.3mol, at equilibrium it will be at 0.3-0.2 = 0.1 mol

[bec]

ah yep yep i did know that actually! thanks!