[rleung]

Hi,

I have the following reaction:

2HI(g) => H2(g) + I2(g)

-d[HI]/dt(1/2) = k[HI]^2 where k = 0.031 1/(M*min)

If I know that at t=20 min, [HI]=0.0752-M, then

-d[HI]/dt(1/2) = 0.031[0.0752]^2

-d[HI]/dt(1/2) = 1.75 x 10^ -4 M/min

Shouldn't that mean that d[HI]/dt also DECREASES at 2 x 1.75 x 10^ -4 M/min, while d[I2]/dt INCREASES at 1.75 x 10^ -4 M/min?

Thanks.

Ryan

[sdekivit]

if you could explain the notation -d[HI] / dt(1/2) i might be able to help. I've never seen this notation before.

[rleung]

Oh, sorry, -d[HI] / dt(1/2) means the change in molar concentration of [HI] per unit time.  As in...

aA + bB => cC + dD

-d[A]/dt(1/a) = -d/dt(1/b) = d[C]/dt(1/c) = d[D]/dt(1/d]

Ryan

[sdekivit]

ok, but then your notation is not completely correct. But here is another view on the problem:

aA + bB => cC + dD

if we define v = -d[A]/dt then we can say:

v = -d[A]/dt = (-a/b)*d/dt = (a/c)*d[C]/dt = (a/d)*d[D]/dt

The same we can do if we define for example v = -d/dt

Now we have the equation 2HI --> H2 + I2

we define the rate of the reaction as v = -d[HI]/dt = 1,75 x 10^-4 M/min

You have nothing to cahnge for that, since we know the reaction rate at t = 20, because we could measure [HI].

Now we can also say: v = -d[HI]/dt = 2 * d[H2]/dt = 2 * d[I2]/dt

since the reaction rate for HI is twice as large as for I2 and H2 (when 2 mol HI reacts, 1 mol H2 and 1 mol I2 will be formed).

Now you also know the reaction rates for H2 and I2.

[rleung]

Thanks.  I sort of see now.  I think the thing I was confused on was what to use as the "reference" chemical.  For example, I knew that I2 would always have a rate that is half that of HI, but when I found that -d[HI]/dt(1/2) = 1.75 x 10^ -4 M/min, it could go two ways:

1) rate of [HI] consumption is 1.75x10^-4 M/min, while rate of [I2] formation is (1.75x10^-4 M/min)/2

or...

2) rate of [HI] consumption is 2*(1.75x10^-4 M/min), while rate of [I2] formation is 1.75x10^-4 M/min. 

In both of the above cases, rate of [I2] formation would be equal to half of the rate of [HI] consumption, but how do I know which one to go with?  I would think to go with #2 since the original piece of info given was that -d[HI]/dt(1/2) = 1.75 x 10^ -4 M/min.  Accordingly, you can then assume that -d[HI]/dt(1/2) = d[I2]/dt = 1.75 x 10^ -4 M/min, and when you multiple everything by 2, shouldn't you get that -d[HI]/dt = 2d[I2]/dt = 2(1.75 x 10^ -4), and when you solve for each of the differential terms, you should get that -d[HI]dt  = 2(1.75 x 10^ -4), while d[I2]dt = 1.75x10^-4 M/min, no?

Thanks so much.

Ryan

[sdekivit]

it is the first one, since you already know what the rate of reacting of HI is at the moment t = 20. This is calculated by using k * [HI]² . In other words: -d[HI]/dt = 1.75 x 10^-4 M/min.

so at the moment t = 20 we measure [HI] and we calculate the rate of reaction.

Then HI is the 'reference' chemical and thus rate of formation of H₂ and I₂ is half of that of HI

[rleung]

Ah, I see.  Thank you.

[rleung]

Sorry to resurrect this post, but after looking at this problem again, I am confused still.  In the original problem, it says that -(1/2)d[HI] / dt = 0.031[HI]^2, and after finding the concentration of [HI] at t=20, I plugged it into the equation to get 1.75e-4 M/min. 

So, if -(1/2)d[HI] / dt = 1.75e-4 M/min, wouldn't that mean that d[HI] / dt = 2(1.75e-4) M/min, while d[I2] / dt and d[H2] / dt both = 1.75e-4 M/min???

Ryan

[rleung]

Any thoughts?  Sorry to bring up post up again, but looking back at all of your explanations, it still doesn't add up.  You did your calculations assuming that -d[HI]/dt = 1.75 x 10-4 M/min, but in the original problem, it was given that -d[HI]/dt*(1/2) = k[HI]^2, so I don't understand how you can just jump from one to the other.  Sorry for seeming stupid.  I guess I convinced myself earlier I understood it when I really didn't.

Ryan

[Donaldson Tan]

2HI(g) => H2(g) + I2(g)

There is no such thing as a referene molecule. The reaction rate is specified per mole of reaction.

From the stoichiometric ratio, -d[HI]:d[H2]:d[I2] = 2:1:1

hence, v = -(1/2)d[HI]/dt = d[H2]/dt = d[I2]/dt where v is the reaction rate

For example, let the reaction rate be 5 M per minute (ie. v = 5 M/min), so- d[HI]/dt = 10M/min, d[H2]/dt = d[I2]dt = 5M/min

-d[HI]/dt(1/2) = k[HI]^2 where k = 0.031 1/(M*min)

If I know that at t=20 min, [HI]=0.0752-M, then

at t=20min, the rate v = k.[HI]² = 1.75E-4 M/min

-d[HI]/dt = 2 x 1.75E-4 M/min
d[H2]/dt = d[I2]/dt = 1.75E-4 M/min

[sdekivit]

'at t=20min, the rate v = k.[HI]2 = 1.75E-4 M/min'

'-d[HI]/dt = 2 x 1.75E-4 M/min'

this strokes with each other, since the reaction rate is the difference in [HI] in time which is already known ( v = -d[HI]/dt in this case)