Questions: The rate of a certain reaction is given by the following rate law: rate = k[H2]^2[NH3] Use this information to answer the questions below. What is the reaction order in H2? What is the reaction order in NH3? What is overall reaction order? At a certain concentration of H2 and NH3, the initial rate of reaction is 91.0 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to 3 significant digits. The rate of the reaction is measured to be 0.330 M / s when [H2]=1.3 M and [NH3]=1.4 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.

The rate of a certain reaction is given by the following rate law: rate = k[H2]^2[NH3] Use this information to answer the questions below. What is the reaction order in H2? What is the reaction order in NH3? What is overall reaction order? At a certain concentration of H2 and NH3, the initial rate of reaction is 91.0 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to 3 significant digits. The rate of the reaction is measured to be 0.330 M / s when [H2]=1.3 M and [NH3]=1.4 M. Calculate the value of the rate constant. Round your answer to 2 significant digits.
Transcript text: The rate of a certain reaction is given by the following rate law: \[ \text { rate }=\mathrm{k}\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{NH}_{3}\right] \] Use this information to answer the questions below. What is the reaction order in $\mathrm{H}_{2}$ ? What is the reaction order in $\mathrm{NH}_{3}$ ? What is overall reaction order? At a certain concentration of $\mathrm{H}_{2}$ and $\mathrm{NH}_{3}$, the initial rate of reaction is $91.0 \mathrm{M} / \mathrm{s}$. What would the initial rate of the reaction be if the concentration of $\mathrm{H}_{2}$ were doubled? Round your answer to 3 significant digits. The rate of the reaction is measured to be $0.330 \mathrm{M} / \mathrm{s}$ when $\left[\mathrm{H}_{2}\right]=1.3 \mathrm{M}$ and $\left[\mathrm{NH}_{3}\right]=1.4 \mathrm{M}$. Calculate the value of the rate constant. Round your answer to 2 significant digits.
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Solution

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Solution Steps

Step 1: Determine the Reaction Order in \(\mathrm{H}_{2}\)

The rate law is given by:

\[ \text{rate} = k[\mathrm{H}_{2}]^{2}[\mathrm{NH}_{3}] \]

The exponent of \([\mathrm{H}_{2}]\) in the rate law is 2. Therefore, the reaction order with respect to \(\mathrm{H}_{2}\) is 2.

Step 2: Determine the Reaction Order in \(\mathrm{NH}_{3}\)

The exponent of \([\mathrm{NH}_{3}]\) in the rate law is 1. Therefore, the reaction order with respect to \(\mathrm{NH}_{3}\) is 1.

Step 3: Determine the Overall Reaction Order

The overall reaction order is the sum of the exponents in the rate law. Therefore, the overall reaction order is:

\[ 2 + 1 = 3 \]

Step 4: Calculate the New Initial Rate When \([\mathrm{H}_{2}]\) is Doubled

The initial rate of reaction is given as \(91.0 \, \mathrm{M/s}\). If the concentration of \([\mathrm{H}_{2}]\) is doubled, the new rate can be calculated using the rate law:

\[ \text{New rate} = k(2[\mathrm{H}_{2}])^{2}[\mathrm{NH}_{3}] \]

This simplifies to:

\[ \text{New rate} = 4k[\mathrm{H}_{2}]^{2}[\mathrm{NH}_{3}] = 4 \times 91.0 \, \mathrm{M/s} = 364 \, \mathrm{M/s} \]

Final Answer

  • Reaction order in \(\mathrm{H}_{2}\): \(\boxed{2}\)
  • Reaction order in \(\mathrm{NH}_{3}\): \(\boxed{1}\)
  • Overall reaction order: \(\boxed{3}\)
  • New initial rate when \([\mathrm{H}_{2}]\) is doubled: \(\boxed{364 \, \mathrm{M/s}}\)
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