To find the molar concentration of NOBr at equilibrium for the reaction:
[tex]\[ 2 \text{NOBr} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Br}_2 (g) \][/tex]
we will use the given equilibrium constant [tex]\( K_c \)[/tex] and the concentrations of NO and Br[tex]\(_2\)[/tex].
Given:
- [tex]\( K_c = 2.0 \)[/tex]
- [tex]\([ \text{NO} ] = 1.8 \text{ M} \)[/tex]
- [tex]\([ \text{Br}_2 ] = 1.1 \text{ M} \)[/tex]
The equilibrium constant expression for this reaction is:
[tex]\[ K_c = \frac{[\text{NO}]^2 [\text{Br}_2]}{[\text{NOBr}]^2} \][/tex]
We need to solve for [tex]\([\text{NOBr}] \)[/tex]:
[tex]\[ K_c = \frac{[\text{NO}]^2 [\text{Br}_2]}{[\text{NOBr}]^2} \][/tex]
[tex]\[ \Rightarrow [\text{NOBr}]^2 = \frac{[\text{NO}]^2 [\text{Br}_2]}{K_c} \][/tex]
[tex]\[ [\text{NOBr}] = \sqrt{\frac{[\text{NO}]^2 [\text{Br}_2]}{K_c}} \][/tex]
Now plug in the known values:
[tex]\[ [\text{NOBr}] = \sqrt{\frac{{(1.8 \text{ M})}^2 \cdot 1.1 \text{ M}}{2.0}} \][/tex]
Calculate the values inside the square root:
[tex]\[ [\text{NOBr}] = \sqrt{\frac{3.24 \text{ M}^2 \cdot 1.1 \text{ M}}{2.0}} \][/tex]
[tex]\[ [\text{NOBr}] = \sqrt{\frac{3.564 \text{ M}^3}{2.0}} \][/tex]
[tex]\[ [\text{NOBr}] = \sqrt{1.782 \text{ M}^3} \][/tex]
[tex]\[ [\text{NOBr}] = 1.334 \text{ M} \][/tex]
We need to express the answer to two significant figures:
[tex]\[ [\text{NOBr}] = 1.3 \text{ M} \][/tex]
Thus, the molar concentration of NOBr is [tex]\( 1.3 \text{ M} \)[/tex].
