One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.

Short Answer

Expert verified
The rate of the reaction will remain the same.

Step by step solution

01

Understanding Zero-Order Reaction

A zero-order reaction is a reaction whose rate remains constant and is not influenced by changes in the concentration of the reactants. The rate equation for the zero-order reaction is: Rate = k[A]^0 where [A] is the concentration of the reactant, k is the rate constant, and the 0 is the order of the reaction.
02

Applying the Principle to the Reaction

The given reaction is a zero-order reaction: \(2 NH_{3}(g) \longrightarrow N_{2}(g)+3 H_{2}(g)\). According to the principle defined earlier, even if the concentration of ammonia is doubled, it will not affect the rate of reaction.
03

Conclusion

From the above discussion, if the concentration of ammonia is doubled, the rate of reaction will remain constant and not alter.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the area of chemistry that concerns the rates at which chemical reactions occur. It dives into understanding how various factors, like temperature, pressure, and concentration of reactants, affect the speed of a reaction. It also involves the study of reaction mechanisms and the pathway by which a reaction proceeds.

Zero-order reactions fall under the canopy of chemical kinetics, where the rate is independent of the concentration of the reactants. This peculiarity makes zero-order reactions especially interesting because no matter how much you increase or decrease the concentration, the reaction pace stays unwavering. This equilibrium is sustained until one of the reactants is completely consumed, at which point the reaction stops abruptly. Zero-order kinetics are often found in processes that are catalyzed by enzymes or on surfaces where the catalyst is saturated.
Reaction Rate
The reaction rate, simply put, is how swiftly a reaction proceeds. It is typically measured by the change in concentration of reactants or products per unit time. In formulas, it can be represented as the negative of the change in concentration of reactants divided by the change in time (for a disappearance) or the positive of the change in concentration of products divided by the change in time (for an appearance).

In a zero-order reaction, where the reaction rate is measured as Rate = k[A]^0, the superscript '0' notates that the rate of the reaction is not dependent on the concentration of reactant A. In this case, k is a constant known as the rate constant, which is specific to each chemical reaction at a given temperature. The reaction remains at a fixed pace because of a surface process or a catalyst that operates at full capacity and is not affected by how much reactant is present.
Concentration of Reactants
In most chemical reactions, the concentration of reactants is vital in determining the rate at which the reaction proceeds. Altering these concentrations typically speeds up or slows down the reaction accordingly. This relationship is often expressed in the rate law, which indicates how the rate is affected by the concentration of the reactants.

However, this is not the case with zero-order reactions. For these reactions, the rate law shows that the concentration of reactants plays no role in the speed of the reaction. The rate of a zero-order reaction is constant and given by the rate constant, which is not altered by changes in concentration. This exceptional attribute of zero-order kinetics is crucial for students to understand, as it deviates from the more common concentration-dependent reactions they might encounter.

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Most popular questions from this chapter

The following statements about catalysis are not stated as carefully as they might be. What slight modifications would you make in them? (a) A catalyst is a substance that speeds up a chemical reaction but does not take part in the reaction. (b) The function of a catalyst is to lower the activation energy for a chemical reaction.

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

You want to test the following proposed mechanism for the oxidation of HBr. $$\begin{array}{c} \mathrm{HBr}+\mathrm{O}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{HOOBr} \\\ \mathrm{HOOBr}+\mathrm{HBr} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{HOBr} \\\ \mathrm{HOBr}+\mathrm{HBr} \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}+\mathrm{Br}_{2} \end{array}$$ You find that the rate is first order with respect to HBr and to \(\mathrm{O}_{2}\). You cannot detect HOBr among the products. (a) If the proposed mechanism is correct, which must be the rate-determining step? (b) Can you prove the mechanism from these observations? (c) Can you disprove the mechanism from these observations?

For the reaction \(\mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D},\) the rate law is rate of reaction \(=k[\mathrm{A}][\mathrm{B}]\) (a) Show that the following mechanism is consistent with the stoichiometry of the overall reaction and with the rate law. $$\begin{array}{l} \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{I} \quad(\text { slow }) \\ \mathrm{I}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} \quad(\text { fast }) \end{array}$$ (b) Show that the following mechanism is consistent with the stoichiometry of the overall reaction, but not with the rate law. $$\begin{array}{c} 2 \mathrm{B} \stackrel{k_{1}}{\mathrm{k}_{1}} \mathrm{B}_{2} \text { (fast) } \\\ \mathrm{A}+\mathrm{B}_{2} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}+\mathrm{D} \text { (slow) } \end{array}$$

The decomposition of dimethyl ether at \(504^{\circ} \mathrm{C}\) is $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$$ The following data are partial pressures of dimethyl ether (DME) as a function of time: \(t=0\) s, \(P_{\text {DME }}=\) \(312 \mathrm{mmHg} ; 390 \mathrm{s}, 264 \mathrm{mmHg} ; 777 \mathrm{s}, 224 \mathrm{mmHg} ; 1195 \mathrm{s},187 \mathrm{mmHg} ; 3155 \mathrm{s}, 78.5 \mathrm{mmHg}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is the total gas pressure at 390 s? (d) What is the total gas pressure when the reaction has gone to completion? (e) What is the total gas pressure at \(t=1000\) s?

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