An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At equilibrium, the concentrations are [H2] 5.0 M,[N 2] 8.0 M, and [NH3] 4.0 M. What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

[N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

       N2                       3H2                2NH3

I      [N2]₀                     [H2]₀                  0

C     - x                          -3x                 +2x

E     [N2]₀ - x =8      [H2]₀ - 3x =5       2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

Conc. Of H =11

Conc. Of N =10

Explanation:

Provided that the equation for thr reaction is

3H2 + N2 <---> 2NH3

Then

I = 5.0+3x 8.0+x 0

C = -3x -x +2x

E = 5.0 8.0 4.0

But

x=2

So

H2=5.0+3x

5+6=11

N2= 8.0+x

N2=10

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"