Question

- Photon336

Question

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Photon336

388/389

##### 1 Attachment

- Photon336

386/387

##### 1 Attachment

- Photon336

@taramgrant0543664 @Rushwr @sweetburger @Woodward

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Photon336

388 @rushwr?

- Rushwr

thinking

- Photon336

\[Ca _{s} + Cl _{2g} \rightarrow CaCl _{2s}\]

- Rushwr

is it B?

- Rushwr

cuz the reaction is a formation reaction hence it is an exothermic reaction. SO delta H is negative

- Rushwr

\[\Delta S _{surr} =\frac{ -\Delta H }{ T }\]

- Photon336

\[\Delta G = \Delta H - T \Delta S \]
so for our reaction the entropy would have to go down.
so the sign TDeltaS would be positive.
that means that for our reaction, which we know is spontaneous because we form our reactants. Delta G < 0
if we put the two together.
1. Delta G <0
2. Entropy Delta S < 0
3. TdeltaS > 0
THIS means that T delta S would be > 0 positive.
and Delta G < 0
therefore the enthalpy Delta H would have to be large and negative, to offset the positive entropy and that would make delta G < 0

- Photon336

sorry form our products***

- Rushwr

389. A Cuz when they had more reactants in to the system the system tend to attain the equilibrium by making more products. So the euilibrium points shifts to the left forming more products.

- Photon336

\[\Delta G = \Delta H - T \Delta S \]
True entropy is \[\Delta S = \frac{ -\Delta H }{ T }\]
because we end up with solid calcium chloride (s) we know overall the entropy goes down.
so in that equation. delta S < 0
\[\Delta S < 0 \]
\[\Delta H < 0 \]

- Photon336

389 is so tricky the first time I did it i got it wrong

- Photon336

@taramgrant0543664 thoughts for 389?

- Rushwr

what's the answer given for 389. I explained one !

- taramgrant0543664

So I was thinking at first A until I saw that it was in solid form solids don't change rate so I was considering c

- Photon336

@Rushwr it's C no change

- Rushwr

oh yes it doesn't count for the K values yes. u are right!

- Photon336

equilibrium expressions, the K value(s) i think it's pure liquids and solids aren't included.

- Rushwr

I didn't see the states properly ! Thank you @taramgrant0543664 and @Photon336

- taramgrant0543664

For it to be counted it's either in gas or aqueous form

- Rushwr

@taramgrant0543664 I love you!

- taramgrant0543664

Definitely watch those states that's the thing that always gets me lol

- Photon336

\[Keq = [Cl _{2}] \]
\[K _{eq} = \frac{ [C][D] }{ [A][B] }\]

- Photon336

yeah... that one got me too..

- taramgrant0543664

Was the answer for 388 said? I might have missed that I know I missed part of your conversation since I got tagged in the math section

- Photon336

it remains the same because the Keq depends only on the concentration of the gas chlorine.

- Photon336

so it's C

- taramgrant0543664

Right I was thinking that but thought I would double check as sometimes I am completely off on them lol what's the next one?

- Photon336

387. A. because like replacing a strong acid with a weak one in that reaction would cause the temperature change to rise to a smaller degree, and the enthalpy change would be less.

- Photon336

I just gave an answer for 387, so all that's left is 386

- Photon336

I will post one more set of questions then I g2g

- Rushwr

okai !

- Rushwr

Btw Thanks alot @Photon336

- Photon336

386 I missed that one

- taramgrant0543664

These are kind of fun to do @Photon336!!
As for 386 I was considering D as my option possibly?

- Rushwr

IKR this is awesome ! @taramgrant0543664 @Photon336

- Photon336

Yeah it's D.

- Photon336

ok Will post one more.

Looking for something else?

Not the answer you are looking for? Search for more explanations.