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Chapter 14: Problem 1

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

Short Answer

Expert verified
The rate of reaction is \(3.1 \times 10^{-4}\) M/s, the rate of disappearance of B is \(-3.1 \times 10^{-4}\) M/s, and the rate of formation of D is \(9.3 \times 10^{-4}\) M/s.

Step by step solution

01

Determine the Rate of Reaction

The rate of reaction is the rate of disappearance of any one of the reactants or the rate of appearance of any one of the products. Given the rate of disappearance of reactant A (-6.2 x 10^-4 M/s), the rate of the reaction is simply this rate divided by the stoichiometric coefficient of A in the balanced equation, i.e., 2. The equation for determining this is \(-\frac{1}{2}\) \(\times Rate_{a}\) which gives us - (-6.2 x 10^-4) / 2 = 3.1 x 10^-4 M/s.
02

Determine the Rate of Disappearance of B

The rate of disappearance of B is the rate of reaction multiplied by the stoichiometric coefficient of B. As there's no coefficient for B presented in the balanced equation, it means that coefficient for B is 1. The equation for determining the disappearance rate of B is \(-Rate_{reaction}\).
03

Determine the Rate of Formation of D

The rate at which D is formed is equal to the rate of reaction multiplied by the stoichiometric coefficient of D in the balanced equation. The stoichiometric coefficient of D is 3, so we can use the equation \(Rate_{reaction}\) \(\times 3\) to solve.

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Most popular questions from this chapter

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

Explain why (a) A reaction rate cannot be calculated from the collision frequency alone. (b) The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency increases much more slowly. (c) The addition of a catalyst to a reaction mixture can have such a pronounced effect on the rate of a reaction, even if the temperature is held constant.

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?

The reaction \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) has the rate law: rate of reaction \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] .\) Propose a twostep mechanism for this reaction consisting of a fast reversible first step, followed by a slow step.

For the disproportionation of \(p\)-toluenesulfinic acid, $$3 \mathrm{ArSO}_{2} \mathrm{H} \longrightarrow \mathrm{ArSO}_{2} \mathrm{SAr}+\mathrm{ArSO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O}$$ (where \(\mathrm{Ar}=p-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4}-\) ), the following data were obtained: \(t=0 \min ,[\mathrm{ArSO}_{2} \mathrm{H}]=0.100 \mathrm{M} ; 15 \mathrm{min}, 0.0863 \mathrm{M} ; 30 \mathrm{min}, 0.0752 \mathrm{M} ; 45 \mathrm{min}, 0.0640 \mathrm{M} ; 60 \mathrm{min}, 0.0568 \mathrm{M} ; 120 \mathrm{min}, 0.0387 \mathrm{M} ; 180 \mathrm{min}, 0.0297 \mathrm{M}; 300 \mathrm{min}, 0.0196 \mathrm{M}.\) (a) Show that this reaction is second order. (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0500 \mathrm{M} ?\) (d) At what time would \(\left(\mathrm{ArSO}_{2} \mathrm{H}\right)=0.0250 \mathrm{M} ?\) (e) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0350 \mathrm{M} ?\)
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