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Give three related expressions for the rate of the reaction

\({N_2}\;\left( g \right)\; + \;3{H_2}\left( g \right)\;\; \to \;2N{H_3}\left( g \right)\)

assuming that the concentrations of any intermediates are constant and that the volume of the reaction vessel does not change.

Short Answer

Expert verified

The rate expression is given as the below:

\(rate\; = \; - \dfrac{{d\left( {{N_2}} \right)}}{{dt}}\; = \; - \dfrac{1}{3}\dfrac{{d\left( {{H_2}} \right)}}{{dt}}\; = \;\dfrac{1}{2}\dfrac{{d\left( {N{H_3}} \right)}}{{dt}}\)

Step by step solution

01

STEP-1:   Finding the rate expression for the given equation

For a chemical reaction rate expression can be defined as rate of change of concendration of every molecule or species with respect to time in an equation.

As the concendration of reactents decreases with time the reactents are represented by negative sign,whereas positive sign is represented in the products side as the concendration of products increases.

For the reaction

\({N_2}\;\left( g \right)\; + \;3{H_2}\left( g \right)\;\; \to \;2N{H_3}\left( g \right)\)

Rate expression is given as

\(rate\; = \; - \dfrac{{d\left( {{N_2}} \right)}}{{dt}}\; = \; - \dfrac{1}{3}\dfrac{{d\left( {{H_2}} \right)}}{{dt}}\; = \;\dfrac{1}{2}\dfrac{{d\left( {N{H_3}} \right)}}{{dt}}\)

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Most popular questions from this chapter

Write the overall reaction and rate laws that correspond to the following reaction mechanisms. Be sure to eliminate intermediates from the answers.

(a) A+Bโ†’k2k1C+D(Fastequilbrium)C+Eโ†’k2F(Slow)

(b) AโŸทk-1k1B+C(Fastequilbrium)C+DโŸทk-1k1E(Fastequilbrium)Eโ†’k3F(Slow)

The rate constant of the elementary reaction

BrO(g)+NO(g)โ†’Br(g)+NO2(g)

is 1.3ร—1010 Lmol-1 and its equilibrium constant is 5.0 ร— 1010 at this temperature. Calculate the rate constant at 25ยฐC of the elementary reaction

Br(g)+NO2(g)โ†’BrO(g)+NO(g)

The following observations have been made about a certain reacting system: (i) When A, B, and C are mixed at about equal concentrations in a neutral solution, two different products are formed, D and E, with the amount of D about 10 times greater than the amount of E. (ii) If everything is done as in (i) except that a trace of acid is added to the reaction mixture, the same products are formed, except that now the amount of D produced is much smaller than (about 1% of) the amount of E. The acid is not consumed in the reaction. The following mechanism has been proposed to account for some of these observations and others about the order of the reactions:

(1)A+BโŸทk-1k2FRapidequilbrium2C+Fโ†’k2โ€ŠDnegligablereverserate3C+Fโ†’k3โ€ŠEnegligablereverserate

a) Explain what this proposed scheme of reactions implies about the dependence (if any) of the rate of formation of D on the concentrations of A, of B, and of C. What about the dependence (if any) of the rate of formation of E on these same concentrations?

b) What can you say about the relative magnitudes of k2and k3?

c) What explanation can you give for observation (ii) in view of your answer to (b)?

HCl reacts with propene (CH3CHCH2) in the gas phase according to the overall reaction.

HCl + CH3CH CH2 โ†’ CH3CHClCH3

The experimental rate expression is

rate = k[ HCl ]3 [ CH3CHCH2 ]

Which, if any, of the following mechanisms are consistent with the observed rate expression?

(a) HCl + HCl โ‡„ H + HCl2 (Fast)

H + CH3CH CH2 โ†’ CH3CH CH3(Slow)

HCl2 + CH3CH CH3 โ†’ CH3CHClCH3 + HCl (Fast)

(b) HCl + HCl โ‡„ H2Cl2

HCl + CH3CHCH2 โ†’ CH3CHCl CH3*(Slow)

CH3CHCl CH3*+ H2Cl2โ†’ CH3CHCl CH3+ 2HCl

(C) HCl + CH3CH CH2 โ†’ H +CH3CHClCH2 (Fast equilbrium)

H + HCl โ‡„ H2Cl (Fast equilbrium)

H2Cl + CH3CHCl CH2 โ†’ HCl + CH3CHCl CH3 (Slow) ,

At 600 K, the rate constant for the first-order decomposition of nitroethane CH3CH2NO2โ†’C2H4+HNO2is 1.9 ร— 10-4s-1 . A sample of CH3CH2NO2

is heated to 600K, at which point its initial partial pressure is measured to be 0.078 atm. Calculate its partial pressure after 3.0hours.

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