Question

i can calculate the reaction quotient, q and determine which direction a reaction will proceed to reach equilibrium.

1. for the synthesis of ammonia gas at 875k in a closed vessel from nitrogen and hydrogen gases, the equilibrium constant ( k_{\text{eq}} ) is ( 6.0 \times 10^{-2} ). if the measured concentrations are ( \text{nh}_3 = 2.00 \times 10^{-4} \text{mol/l} ); ( \text{n}_2 = 1.50 \times 10^{-5} \text{mol/l} ); and ( \text{h}_2 = 0.354 \text{mol/l} ), is the system at equilibrium? if not, which way will the system shift to reach equilibrium?

( \text{n}_2 \text{(g)} + 3 \text{h}_2 \text{(g)} leftrightarrow 2 \text{nh}_3 \text{(g)} )
youve got it!
next steps:

Explanation:

Step1: Recall Reaction Quotient Formula

For reaction \( \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \), \( Q = \frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} \).

Step2: Substitute Concentrations

Plug in \( [\ce{NH3}] = 2.00 \times 10^{-4} \, \text{mol/L} \), \( [\ce{N2}] = 1.50 \times 10^{-5} \, \text{mol/L} \), \( [\ce{H2}] = 0.354 \, \text{mol/L} \):
\( Q = \frac{(2.00 \times 10^{-4})^2}{(1.50 \times 10^{-5})(0.354)^3} \).

Step3: Calculate Numerator and Denominator

Numerator: \( (2.00 \times 10^{-4})^2 = 4.00 \times 10^{-8} \).
Denominator: \( (1.50 \times 10^{-5})(0.354)^3 \approx (1.50 \times 10^{-5})(0.0443) \approx 6.65 \times 10^{-7} \).

Step4: Compute Q

\( Q = \frac{4.00 \times 10^{-8}}{6.65 \times 10^{-7}} \approx 0.0602 \).

Step5: Compare Q and \( K_{eq} \)

\( K_{eq} = 6.0 \times 10^{-2} = 0.060 \).
Since \( Q \approx 0.0602 \approx K_{eq} \), the system is at equilibrium.

Answer:

The system is at equilibrium.