Calculate the concentration (in mol/L) of the products of the following equilibrium when the equilibrium constant is 19.0 and the equilibrium concentration of NOBr2(g) is 2.69 mol/L.
NOBr2(g) → NO(g) + Br2(g)
5 years ago
Answered By Leonardo F
We can start this exercise by writing the expression for the equilibrium constant K. Since all the coefficients in the chemical equation are 1:
Since the stoichiometric proportion of NO and $Br_2$Br2 is 1:1 in the products, they are in the same physical state, and the reaction is at equilibrium, they will have the same constant concentration:
$\left[NO\right]=\left[Br_2\right]$[NO]=[Br2] at equilibrium
Substituting this result in the previous equation:
5 years ago
Answered By Leonardo F
We can start this exercise by writing the expression for the equilibrium constant K. Since all the coefficients in the chemical equation are 1:
$K=\frac{\left[NO\right]\left[Br_2\right]}{\left[NOBr_2\right]}$K=[NO][Br2][NOBr2]
Putting the values given in the exercise:
$19.0=\frac{\left[NO\right]\left[Br_2\right]}{2.69}$19.0=[NO][Br2]2.69
$\left[NO\right]\left[Br_2\right]=19\times2.69=51.11$[NO][Br2]=19×2.69=51.11
Since the stoichiometric proportion of NO and $Br_2$Br2 is 1:1 in the products, they are in the same physical state, and the reaction is at equilibrium, they will have the same constant concentration:
$\left[NO\right]=\left[Br_2\right]$[NO]=[Br2] at equilibrium
Substituting this result in the previous equation:
$\left[NO\right]^2=51.11$[NO]2=51.11
$\left[NO\right]=\sqrt{51.11}=\frac{7.15mol}{L}$[NO]=√51.11=7.15molL
Since the $Br_2$Br2 will have the same concentration, $\left[Br_2\right]=\frac{7.15mol}{L}$[Br2]=7.15molL