Question

Calculate K at 298 K for the following reactions. Delta Gf kj/mol = NOg...... 86.60 NO2g........... 51 O2.....0 Part A : NO(g) + 1/2 O2(g)......>> NO2(g) ..(.I know the answer is 1.7 x 10^6) Part B) Use the equilibrium constant ( K= 3.89 x 10^-34 at 298K ) to calculate Delta G (in kJ) at 298 for the following reaction: 2HCL (g)....> H2 (g) + CL2 (g) answer delta G in kJ and in scientific notation please.... NOTE:... the answer to part a is correct, the answer to part b is NOT.-1.90 x 10^2 it said HInt: For any state function (G, H, or S) the change in the state function, F, for a reaction can be calculated from Delta F rxn= (sideways M)n x delta F (products) minus (sideways M) n x delta F (reactants), when n are the stoichiometric coefficients from the balanced chemical equation. Use this expression to calculate Delta G rxn. Use the expression that relates change in fuel energy of a reaction, Delta G rxn, to the equilibrium constant K. I just need part B.....thanks. View comments ...part of the answer to part B is ............10^2 Comments You commented it said ^2 not ^-2 in the answer.......sorry