standard free energy change of -11.70 kJ mol-1. Calculate equilibrium constant for 25 degrees C.

Question

Draw a detailed visual representation of a science laboratory setting. The lab should contain various elements like a microscope, test tubes, scientific calculator, and a chalkboard with an equation. The equation on the board should not be readable, maintaining the 'no text' requirement. Also, include a confused scientist - an African female, pondering over some experiment-result. Place next to her a graph depicting a correlation, again with no readable text or labels.

standard free energy change of -11.70 kJ mol-1. Calculate equilibrium constant for 25 degrees C.

I got 7.772..and it's wrong

I did
-11700J/ (-2.303 x 8.314 x 298)= 2.0505277
and then e^(2.0505277)= 7.77

I don't know what I did wrong

Answer 1

One thing wrong is you used 2.303 which converts ln (base e) to log(base 10). So if you use 2.303 then you should look up 10^2.0505 and not e^2.505. (I said that wrong. Using 2.303 is ok. What isn't proper is to use the 2.303 conversion factor and not follow through with 10^2.505.

Answer 2

thank you!

Answer 3

To calculate the equilibrium constant (K) using the standard free energy change (∆G°), you need to use the equation:

∆G° = -RT ln(K)

First, convert the temperature from Celsius to Kelvin:

T = 25 °C + 273.15 = 298.15 K

Then, rearrange the equation to solve for K:

K = e^(-∆G° / RT)

Now, plug in the values:

∆G° = -11.70 kJ mol^-1 = -11,700 J mol^-1
R = 8.314 J K^-1 mol^-1
T = 298.15 K

Now let's calculate K step by step:

Step 1: Calculate -∆G° / RT

-∆G° / RT = (-11,700 J mol^-1) / (8.314 J K^-1 mol^-1 * 298.15 K) ≈ -0.04716

Step 2: Calculate e^(-∆G° / RT)

e^(-∆G° / RT) ≈ e^(-0.04716) ≈ 0.9548

So, the calculated equilibrium constant (K) is approximately 0.9548.

Please note that the answer you obtained, 7.77, is not correct. Double-check your calculations to ensure you inputted the correct values and applied the correct units.