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.

  1. 👍 0
  2. 👎 0
  3. 👁 170
asked by Breauna
  1. dG = -RTlnK

    1. 👍 0
    2. 👎 0
    posted by DrBob222

Respond to this Question

First Name

Your Response

Similar Questions

  1. 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
  2. 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
  3. Chemistry

    The standard free energy of activation of one reaction is 94.3 kj/mol^-1 (22.54 kcal/mol^-1). The standard free energy of activation for another reaction is 76.4 kJ/mol^-1 (18.26 kcal/mol^-1). Assume temp of 298K and 1 M conc. By

    asked by Tokey on November 10, 2010
  4. Chemistry

    The standard free energy of activation of a reaction A is 88.6 kJ mol–1 (21.2 kcal mol–1) at 298 K. Reaction B is one hundred million times faster than reaction A at the same temperature. The products of each reaction are 10.0

    asked by Anonymous on November 25, 2012
  5. Chemistry

    The standard free energy of activation of a reaction A is 73.4 kJ mol–1 (17.5 kcal mol–1) at 298 K. Reaction B is one million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ

    asked by Breauna on September 8, 2014
  6. chemistry

    The standard free energy of activation of a reaction A is 78.2 kJ mol–1 (18.7 kcal mol–1) at 298 K. Reaction B is ten million times faster than reaction A at the same temperature. The products of each reaction are 10.0 kJ

    asked by Deidra on December 6, 2012
  7. 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
  8. 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
  9. 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
  10. 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

More Similar Questions