16) A scientist conducts an experiment to determine the rate of the following reaction: \[ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \] If the initial concentration of \( \mathrm{H}_{2} \) was 0.150 M and the concentration of \( \mathrm{H}_{2} \) was 0.050 M after 2.10 s , what is the average rate of the reaction? A. \( 0.048 \mathrm{M} / \mathrm{s} \) B. \( 0.016 \mathrm{M} / \mathrm{s} \) C. \( 0.072 \mathrm{M} / \mathrm{s} \) (1). \( 0.032 \mathrm{M} / \mathrm{s} \) \[ \begin{array}{l} \frac{1}{0.050}=\frac{1}{0.150}+2.10 t \\ \frac{40}{3}=2.102 \end{array} \]
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Bonus Knowledge
To determine the average rate of the reaction, we can use the formula for the average rate: \[ \text{Average Rate} = -\frac{\Delta [\text{H}_2]}{\Delta t} \] Here, \(\Delta [\text{H}_2]\) is the change in concentration of hydrogen over time, which can be calculated as: \[ \Delta [\text{H}_2] = [\text{H}_2]_{\text{initial}} - [\text{H}_2]_{\text{final}} = 0.150 \, \text{M} - 0.050 \, \text{M} = 0.100 \, \text{M} \] The time interval \(\Delta t\) is given as \(2.10 \, \text{s}\). Thus, the average rate turns into: \[ \text{Average Rate} = -\frac{0.100 \, \text{M}}{2.10 \, \text{s}} \approx -0.0476 \, \text{M/s} \] The negative sign denotes the decrease in concentration, but since we are interested in the rate, we can express this as \(0.048 \, \text{M/s}\). Therefore, the answer is A. \(0.048 \, \text{M/s}\). When dealing with reaction rates, it’s crucial to remember that different reactions can have varying stoichiometries, which can influence how you interpret the data. Miscalculating the coefficients in the balanced equation can lead to errors in understanding how fast reactants are consumed or products are formed. Another common mistake is forgetting to account for the units properly—always ensure your concentration changes and time values are in compatible units, as this can easily throw off the calculations. Getting familiar with these details can make a big difference in your results!
