Answer
a. $$K_c = \frac{[ NO ]^{ 2 }[ Br_2 ]}{[ NOBr ]^{ 2 }}$$
b. The system is not at equilibrium.
c. The rate of the forward reaction will initially speed up.
d. At equilibrium:
$[NO] \gt 1.0 \space mole/L$ and $[Br_2] \gt 1.0 \space mole/L$
$[NOBr] \lt 1.0 \space mole/L$
Work Step by Step
a. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ NO ]^{ 2 }[ Br_2 ]}{[ NOBr ]^{ 2 }}$$
b.
1. Find the concentrations for each compound in this system.
- 1 mole in 1 Liter of solution: 1 M
$[NOBr] = [NO] = [Br_2]= 1 \space M$
2. Substitute the values and calculate the constant value if the system is at equilibrium:
$$ \frac{( 1.0 )^{ 2 }( 1.0 )}{( 1.0 )^{ 2 }} = 1.0$$
Since $1.0 \neq 2.0$, the system is not at equilibrium
c. Since $1.0 \lt 2.0$, the amount of products must increase to reach equilibrium. Therefore, the forward reaction will initially speed up.
d. More product will be produced, thus: $[NO] \gt 1.0 \space mole/L$ and $[Br_2] \gt 1.0 \space mole/L$.
To do so, some reactants must be consumed. Thus: $[NOBr] \lt 1.0 \space mole/L$
