Question

Consider the following reaction: $$2\mathrm{NO}(g)+2{\mathrm{H}}_2(g)⟶{\mathrm{N}}_2(g)+2{\mathrm{H}}_2\mathrm{O}(g)$$ (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 $\left[{\mathrm{H}}_2\right]=0.015\mathrm{M}?(\mathrm{c})\mathrm{What}$ 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

(a) The rate law for this reaction is given by: $rate=k[\mathrm{NO}{]}^2[{\mathrm{H}}_2{]}^1$. (b) At concentrations of $[\mathrm{NO}]=0.035\mathrm{M}$ and $[{\mathrm{H}}_2]=0.015\mathrm{M}$, the reaction rate at 1000 K is $5.355\mathrm{M}{\mathrm{s}}^{-1}$. (c) When the concentration of $\mathrm{NO}$ is increased to 0.10 M and the concentration of ${\mathrm{H}}_2$ is 0.010 M, the reaction rate at 1000 K is $60.0\mathrm{M}{\mathrm{s}}^{-1}$.

Step-by-step solution

Write the rate law expression

The rate law for this reaction is first order in H2 and second order in NO. The general rate law expression is given by: $$rate=k[A{]}^m[B{]}^n$$In this case, the rate law will be: $$rate=k[\mathrm{NO}{]}^2[{\mathrm{H}}_2{]}^1$$ #b) Calculate reaction rate with concentrations given#

Plug in given values for rate constant and concentrations

Now we have the rate law expression, and we are given the rate constant at 1000 K, which is $6.0\times {10}^4{\mathrm{M}}^{-2}{\mathrm{s}}^{-1}$, along with the concentrations of NO ($0.035M$) and H2 ($0.015M$). Plug these values into the rate law expression: $$rate=(6.0\times {10}^4{\mathrm{M}}^{-2}{\mathrm{s}}^{-1})[(0.035\mathrm{M}){]}^2[(0.015\mathrm{M}){]}^1$$

Simplify the expression to find the rate

Now, calculate the reaction rate by multiplying the values and their respective units: $$rate=(6.0\times {10}^4{\mathrm{M}}^{-2}{\mathrm{s}}^{-1})({0.035}^2{\mathrm{M}}^2)(0.015\mathrm{M})$$ $$rate=6.0\times {10}^4{M}^{-2}{s}^{-1}\times {0.035}^2{M}^2\times 0.015M$$ $$rate=5.355M{s}^{-1}$$ #c) Calculate new reaction rate with new concentrations given #

Plug in new values for concentrations

Now, we need to calculate the new reaction rate when the concentration of NO is increased to $0.10M$, while the concentration of H2 is $0.010M$ at 1000 K. Plug these values into the rate law expression: $$rate=(6.0\times {10}^4{\mathrm{M}}^{-2}{\mathrm{s}}^{-1})[(0.10\mathrm{M}){]}^2[(0.010\mathrm{M}){]}^1$$

Simplify the expression to find the new reaction rate

Now, calculate the new reaction rate by multiplying the values and their respective units: $$rate=6.0\times {10}^4{M}^{-2}{s}^{-1}\times {0.10}^2{M}^2\times 0.010M$$ $$rate=60.0M{s}^{-1}$$

Key concepts

Rate Law

The rate law is an essential concept in chemical kinetics. It describes the relationship between the rate of a chemical reaction and the concentration of the reactants. In a reaction, the rate law can be expressed in general form as $rate=k[A{]}^m[B{]}^n$, where $k$ is the rate constant, $[A]$ and $[B]$ are the concentrations of the reactants, and $m$ and $n$ are the orders of the reaction with respect to each reactant. For the given exercise, which involves the reaction $2\mathrm{NO}(g)+2{\mathrm{H}}_2(g)⟶{\mathrm{N}}_2(g)+2{\mathrm{H}}_2\mathrm{O}(g)$, the rate law is first order in ${\mathrm{H}}_2$ and second order in $\mathrm{NO}$. Thus, the rate law expression becomes $rate=k[\mathrm{NO}{]}^2[{\mathrm{H}}_2{]}^1$. This informs us how changes in the concentrations of $\mathrm{NO}$ and ${\mathrm{H}}_2$ will affect the reaction rate.

Reaction Rate

The reaction rate refers to the speed at which reactants are converted into products in a chemical reaction. Understanding how to calculate the reaction rate is critical in chemical kinetics. In our exercise, we use the rate law to determine how fast the reaction progresses under given conditions. By plugging the concentrations of $\mathrm{NO}$ and ${\mathrm{H}}_2$ into the rate law, we can calculate the reaction rate.

• With $\mathrm{NO}$ at 0.035 M and ${\mathrm{H}}_2$ at 0.015 M, we find that the reaction rate is 5.355 M\,s^{-1}
• When $\mathrm{NO}$ is increased to 0.10 M and ${\mathrm{H}}_2$ is 0.010 M, the reaction rate changes to 60.0 M\,s^{-1}

This clearly illustrates how variations in reactant concentrations impact the reaction rate.

Rate Constant

The rate constant, symbolized as $k$, is a crucial factor in the rate law equation. It provides the proportionality between the reaction rate and the concentrations of reactants raised to their respective orders. The value of $k$ is specific to a particular reaction at a certain temperature, illustrating its dependence on conditions like temperature. In this exercise, the rate constant $k$ is given as $6.0\times {10}^4{\mathrm{M}}^{-2}{\mathrm{s}}^{-1}$ at 1000 K. The units of the rate constant can vary depending on the reaction order, reflecting the need to multiply the concentration terms to yield a rate with units of concentration per time. The given value of $k$ allows us to compute the reaction rate by applying it in conjunction with the reactant concentrations in the rate law expression.

Order of Reaction

The order of a reaction with respect to a particular reactant denotes the power to which the concentration of that reactant is raised in the rate law. The overall order of a reaction is the sum of the orders with respect to each reactant. For the reaction provided in the task, it is second order overall because it is first order in ${\mathrm{H}}_2$ and second order in $\mathrm{NO}$.

• First Order: Indicates a direct proportionality, i.e., if the concentration doubles, so does the reaction rate.
• Second Order: The rate is proportional to the square of the concentration; doubling the concentration quadruples the rate.

Knowing the order of a reaction helps predict how changing concentrations will alter the reaction speed, which is vital for both experimental kinetics and industrial application. In this exercise, understanding that the reaction is more sensitive to changes in $[\mathrm{NO}]$ due to its higher order, explains why an increase in $[\mathrm{NO}]$ causes a significant increase in the reaction rate.