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Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M}\), while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

Short Answer

Expert verified
(a) The rate law for this reaction is given by Rate = k [NO]^2 [H鈧俔. (b) When [NO]=0.035 M and [H鈧俔=0.015 M, the reaction rate is 2.205 脳 10鈦宦 M s鈦宦. (c) When [NO]=0.10 M and [H鈧俔=0.010 M, the reaction rate is 6.0 脳 10鈦宦 M s鈦宦.

Step by step solution

01

Part (a) - Writing the Rate Law

Given that the reaction is first order in H鈧 and second order in NO, we can write the rate law by using the following equation: Rate = k [NO]^(order in NO) [H鈧俔^(order in H鈧) where k is the rate constant, the order in NO is 2, and the order in H鈧 is 1. So, the rate law is: Rate = k [NO]^2 [H鈧俔
02

Part (b) - Calculating the Reaction Rate for Given Concentrations

With the rate constant k = \(6.0 \times 10^4 M^{-2} s^{-1}\), and the given concentrations of NO and H鈧, we can determine the rate using the rate law equation: Rate = k [NO]^2 [H鈧俔 Rate = (6.0 脳 10^4 M鈦宦 s鈦宦)(0.035 M)^2(0.015 M) Performing the calculation, we get: Rate = 2.205 脳 10鈦宦 M s鈦宦
03

Part (c) - Calculating the Reaction Rate for Different Concentrations

Now we need to calculate the reaction rate at 1000 K when the concentration of NO is increased to 0.10 M, and the concentration of H鈧 is 0.010 M. We can use the rate law equation again: Rate = k [NO]^2 [H鈧俔 Rate = (6.0 脳 10^4 M鈦宦 s鈦宦)(0.10 M)^2(0.010 M) Performing the calculation, we get: Rate = 6.0 脳 10鈦宦 M s鈦宦

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Reaction Order
In chemistry, the term 'reaction order' refers to the power to which the concentration of a reactant is raised in the rate law expression. It indicates how the rate of a chemical reaction is affected by the concentration of the reactants.

For instance, for a reaction where the rate law is expressed as \( Rate = k [A]^m [B]^n \), 'm' and 'n' represent the reaction orders with respect to reactants \( A \) and \( B \), respectively. The overall reaction order is the sum of these individual orders, in this case, \( m+n \).

Understanding reaction orders is essential for predicting how changes in concentrations impact the speed of the reaction. A first-order reaction implies that the rate is directly proportional to the concentration of one reactant. Meanwhile, a second-order reaction means that the rate is proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants.
Determining the Rate Constant
The rate constant, symbolized by 'k', is a proportionality constant in the rate law of a chemical reaction. It is a crucial value that reveals the intrinsic speed of a reaction under certain conditions without any concentration effects factored in.

The units of the rate constant vary depending on the overall reaction order. For example, a second-order reaction will have a rate constant with the unit \( \text{M}^{-1}\text{s}^{-1} \), indicating that the rate is affected by the concentration squared. The exercise mentioned a rate constant \( k = 6.0 \times 10^{4} \text{M}^{-2}\text{s}^{-1} \) at 1000 K, signifying that it relates to a third-order reaction: second-order in NO and first-order in \( H_2 \).

The rate constant depends on temperature and the specific reaction but is independent of the reactant concentrations. Calculating or experimentally determining it is essential for understanding the kinetics of a reaction.
Calculating Reaction Rates
Reaction rate calculation is applying the rate law and the rate constant to figure out how fast a reaction occurs under certain concentrations of reactants. It's a matter of plugging values into the rate equation and computing the result.

In the given exercise, once we have the concentration of reactants NO and \( H_2 \), and the rate constant \( k \), we use the equation \( \text{Rate} = k [\text{NO}]^2 [\text{H}_2] \) to calculate the rate. The rate law is crucial because it quantifies the relationship between reactant concentrations and the reaction's velocity.

Remember that for accurate calculations, one must ensure that the units match and that concentration values are in molarity (M). By knowing the rate, we can predict how altering concentrations or changing the temperature (which affects 'k') will influence the speed of the reaction, allowing for better control and optimization in chemical processes and industries.

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Most popular questions from this chapter

The following data were measured for the reaction $\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g)$ \begin{tabular}{lccl} \hline Experiment & {\(\left[\mathrm{BF}_{3}\right](M)\)} & {\(\left[\mathrm{NH}_{3}\right](M)\)} & Initial Rate \((M / \mathrm{s})\) \\ \hline 1 & \(0.250\) & \(0.250\) & \(0.2130\) \\ 2 & \(0.250\) & \(0.125\) & \(0.1065\) \\ 3 & \(0.200\) & \(0.100\) & \(0.0682\) \\ 4 & \(0.350\) & \(0.100\) & \(0.1193\) \\ 5 & \(0.175\) & \(0.100\) & \(0.0596\) \\ \hline \end{tabular} (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) What is the value of the rate constant for the reaction? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

Explain why rate laws generally cannot be written from balanced equations. Under what circumstance is the rate law related directly to the balanced equation for a reaction?

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after \(5.12 \mathrm{~s}\) assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ?

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