Question

How does deltaG° relate to K for a reaction?

A. When deltaG° > 0, K < 1 and products are favored.
B. When deltaG° < 0, K > 1 and products are favored.
C. When deltaG° < 0, K < 1 and reactants are favored.
D. When deltaG° > 0, K > 1 and reactants are favored.

Answer 1

There's an equation that links \(\Delta G\) and \(K\), have you got it?

Answer 2

No? can you give it to me?

Answer 3

\[\Delta G = -R*T*ln(K_{EQ})\]

Answer 4

Is the answer c?

Answer 5

Also, can someone tell me what it means to bump a question?

Answer 6

And to close a question?

Answer 7

it's not C.

You need to know what a negative \(\Delta G\) \(means\) for a reaction, and how to move the math around so that if you know that \(\Delta G\) is negative, what the value of \(K\) will be.

A negative \(\Delta G\) means that a reaction is spontaneous, and moves towards products. So that eliminates 2 options right away, doesn't it?

Answer 8

does it eliminate b too?

Answer 9

it does not eliminate B. You need to decide whether the first statement is true, about when \(\Delta G\) is negative, what that means for \(K\).


I'll use choice A as an example

Answer 10

By rearranging the equation to solve for \(K\), I get \[K = e^{\frac{-\Delta G}{RT}}\]

If \(\Delta G\) is positive, as in choice A, what does that \(guarantee\) about the sign of \(K\)?

It guarantees that it is going to be negative.

So far that statement is algebraically correct.

The SECOND part of the problem asks what \(direction\) the reaction favors under those conditions. Since we've made \(\Delta G\) positive, that means the reaction WON'T move forward, and the products \(aren't\) favored.

Choice A is incorrect

Answer 11

so therefore it is either b or d?

Answer 12

i think b...

Answer 13

it is B