The reaction N2(g)+3H2(g)<-->2NH3(g) begins with 1.15 M N2 and 1.36 M H2. At equilibrium the concentration of NH3 is 0.78 M.

Determine the extent of reaction.

I don't know what you mean by extent of reaction.

I think it is the percent yield. the answer is 86.029411764706 but i'm not sure how

.......N2 + 3H2 ==> 2NH3

I.....1.15..1.36......0
C......-x....-3x.....2x
E....1.15-x..1.36-3x..2x

So if 2x = 0.78 then x = 0.39
The way they get 86.02% is this.
(3x)/1.36]*100 = 86.029 but that makes no sense to me.
I Googled "extent of reaction" and came up with several sites. Here is one such site.
http://en.wikipedia.org/wiki/Extent_of_reaction

I tried that with no luck.

hmm ok thanks though!

To determine the extent of reaction, we need to compare the change in concentration of reactants and products from the initial state to the equilibrium state.

The balanced equation for the reaction is:
N2(g) + 3H2(g) ↔ 2NH3(g)

Given concentrations:
Initial concentration of N2 (ci_N2) = 1.15 M
Initial concentration of H2 (ci_H2) = 1.36 M
Equilibrium concentration of NH3 (ce_NH3) = 0.78 M

The stoichiometric coefficients in the balanced equation indicate that:
For every 1 mol of N2 reacted, 2 mol of NH3 are formed.
And for every 3 mol of H2 reacted, 2 mol of NH3 are formed.

Let the extent of reaction be denoted as 'x' (in mol).

From the balanced equation, we can write the expression for the equilibrium concentrations in terms of the extent of reaction:
ce_NH3 = 2x
ce_N2 = ci_N2 - x
ce_H2 = ci_H2 - 3x

Substituting the given values into the equations, we get:
0.78 M = 2x
1.15 M - x = 0.78 M
1.36 M - 3x = 0.78 M

Solving these equations, we find:
x = 0.39 M

The extent of reaction is 0.39 M, which means that 0.39 moles of N2 have reacted to form 0.78 moles of NH3.