QuestionJuly 24, 2025

Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0026Mcdot s^-1 2NH_(3)(g)arrow N_(2)(g)+3H_(2)(g) Suppose a 300. mL flask is charged under these conditions with 450.mmol of ammonia. How much is left 100 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol,if necessary, and round it to the correct number of significant digits. square

Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0026Mcdot s^-1 2NH_(3)(g)arrow N_(2)(g)+3H_(2)(g) Suppose a 300. mL flask is charged under these conditions with 450.mmol of ammonia. How much is left 100 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol,if necessary, and round it to the correct number of significant digits. square
Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0026Mcdot s^-1 2NH_(3)(g)arrow N_(2)(g)+3H_(2)(g) Suppose a 300. mL flask is charged under these conditions with 450.mmol of ammonia. How much is left 100 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol,if necessary, and round it to the correct number of significant digits. square

Solution
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Answer

1.24 M Explanation 1. Convert initial moles to molarity Initial moles of NH_3 = 450 mmol = 0.450 mol. Volume = 300 mL = 0.300 L. Molarity = \frac{0.450 \text{ mol}}{0.300 \text{ L}} = 1.5 \text{ M}. 2. Apply zero-order kinetics formula **Zero-order reaction formula**: [A]_t = [A]_0 - kt. Here, k = 0.0026 \text{ M}\cdot s^{-1}, t = 100 \text{ s}, and [A]_0 = 1.5 \text{ M}. 3. Calculate concentration after 100 seconds [NH_3]_t = 1.5 \text{ M} - (0.0026 \text{ M}\cdot s^{-1} \times 100 \text{ s}) = 1.5 \text{ M} - 0.26 \text{ M} = 1.24 \text{ M}.

Explanation

1. Convert initial moles to molarity<br /> Initial moles of $NH_3$ = 450 mmol = 0.450 mol. Volume = 300 mL = 0.300 L. Molarity = $\frac{0.450 \text{ mol}}{0.300 \text{ L}} = 1.5 \text{ M}$.<br />2. Apply zero-order kinetics formula<br /> **Zero-order reaction formula**: $[A]_t = [A]_0 - kt$. Here, $k = 0.0026 \text{ M}\cdot s^{-1}$, $t = 100 \text{ s}$, and $[A]_0 = 1.5 \text{ M}$.<br />3. Calculate concentration after 100 seconds<br /> $[NH_3]_t = 1.5 \text{ M} - (0.0026 \text{ M}\cdot s^{-1} \times 100 \text{ s}) = 1.5 \text{ M} - 0.26 \text{ M} = 1.24 \text{ M}$.
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