chem

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

  1. 👍 0
  2. 👎 0
  3. 👁 89
asked by Yoo
  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.

    1. 👍 0
    2. 👎 0
    posted by DrBob222
  2. thank you!

    1. 👍 0
    2. 👎 0
    posted by Yoo

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    A reaction has a standard free-energy change of –10.30 kJ mol–1 (–2.462 kcal mol–1). Calculate the equilibrium constant at 25ºC.? This is what I've done so far.. ∆Gº= -2.3 RT + log (K) -10.31 kJ = -2.3 (.0083145)(298)

    asked by Angely R. on March 18, 2014
  2. Chemistry

    A reaction has a standard free-energy change of -11.80 kJ/mol (-2.820 kcal/mol). Calculate the equilibrium constant for the reaction at 25 degrees C. Please help!

    asked by Tanisha on November 9, 2011
  3. chemistry

    C2H2(g) + 2 H2(g)--> C2H6(g) Substance So (J/mol∙K) ∆Hºf (kJ/mol) C2H2(g 200.9 226.7 H2(g) 130.7 0 C2H6(g) -- -84.7 Bond Bond Energy (kJ/mol) C-C 347 C=C 611 C-H 414 H-H 436 If the value of the standard entropy change, ∆Sº

    asked by Brenda on January 13, 2011
  4. Chemistry

    C2H2(g) + 2 H2(g)--> C2H6(g) Substance So (J/mol∙K) ∆Hºf (kJ/mol) C2H2(g 200.9 226.7 H2(g) 130.7 0 C2H6(g) -- -84.7 Bond Bond Energy (kJ/mol) C-C 347 C=C 611 C-H 414 H-H 436 If the value of the standard entropy change, ∆Sº

    asked by Brenda on January 11, 2011
  5. Chemistry

    C2H2(g) + 2 H2(g)--> C2H6(g) Substance So (J/mol∙K) ∆Hºf (kJ/mol) C2H2(g 200.9 226.7 H2(g) 130.7 0 C2H6(g) -- -84.7 Bond Bond Energy (kJ/mol) C-C 347 C=C 611 C-H 414 H-H 436 If the value of the standard entropy change, ∆Sº

    asked by Brenda on January 12, 2011
  6. Chemistry

    A reaction has a standard free-energy change of –12.40 kJ mol–1 (–2.964 kcal mol–1). Calculate the equilibrium constant for the reaction at 25 °C.

    asked by Anonymous on November 25, 2012
  7. Chemistry

    A reaction has a standard free-energy change of –12.40 kJ mol–1 (–2.964 kcal mol–1). Calculate the equilibrium constant for the reaction at 25 °C.

    asked by Breauna on September 8, 2014
  8. Science

    A reaction has a standard free-energy change of –11.70 kJ mol–1 (–2.796 kcal mol–1). Calculate the equilibrium constant for the reaction at 25 °C. Im so lost please help

    asked by Roberto on March 20, 2017
  9. physics

    1. The equilibrium constant for a certain system is 50 at 25 celsius. Calculate the free energy change 2. When the fee energy change: a) =0, the position of the equilibrium, temperature * change in entropy, free energy change at

    asked by POPPY on April 8, 2019
  10. Chemistry

    C2H2(g) + 2 H2(g)--> C2H6(g) Information about the substances involved in the reaction represented above is summarized in the following tables. Substance/ So (J/mol∙K) /∆Hºf (kJ/mol) C2H2(g) / 200.9 / 226.7 H2(g) / 130.7 / 0

    asked by Brenda on January 25, 2011

More Similar Questions