Part A Consider the second-order reaction: 2HI(g)H2(g)+I2(g) Rate law: k[H]^2 k=

Question

Part A

Consider the second-order reaction:

2HI(g)H2(g)+I2(g)

Rate law: k[H]^2

k= 6.4*10^-9 (mol*s) at 500 K

Initial rate = 1.6 * 10^-7 mol (l*s)

What will be the concentration of HI after t = 3.65×1010 s ([HI]t) for a reaction starting under this condition?

Part B

In a study of the decomposition of the compound X via the reaction

X(g)Y(g)+Z(g)

the following concentration-time data were collected:

Given that the rate constant for the decomposition of hypothetical compound X from part A is 1.60 M1min1, calculate the concentration of X after 14.0 min .

Part C

The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)2NO2(g):

What is the rate of disappearance of NO when [NO]= 6.15×102 M and [O2]=1.75×102 M ?

Part D

What is the rate of disappearance of O2 at the concentrations given in part (c)?

Part E

A certain reaction with an activation energy of 175 kJ/mol was run at 515 K and again at 535 K . What is the ratio of f at the higher temperature to f at the lower temperature?

Use f=e^Ea/(RT)

Time (min) [X](M) 0 0.467 1 0.267 2 0.187 3 0.144 4 0.117 5 0.099 6 0.085 7 0.075

Explanation / Answer

Part A

given : rate=K[HI]^2

If initial rate=1.6*10^-7 mol/ L s

then [HI]o=initial concentration= (initial rate/K)^1/2= sqrt (1.6*10^-7 mol/ L s/6.4*10^-9 L/mol S)=5.0 mol/L

using rate equation for second order,

Kt=1/[HI]o-1/[HI]t

(6.4*10^-9 L/mol S) * 3.65*10^10 s=1/5.0 mol/L-1/[HI]t

solving this,

233.6 mol-1 L=0.2 mol-1 L-1/[HI]t

[HI]t=4.3*10^-3 mol/L

part B

K=1.60M-1 min-1, unit suggests that it is a second order reaction

[X]o=initial conc=0.467,[X]t =conc after time t

Kt=1/[X]o-1/[X]t

1.60M-1 min-1,* 14.0 min=1/0.467 M-1/[X]t

[X]t=4.9*10^-2 M

part C

Rate=[NO]^2 [O2]=-1/2 d[NO]/dt=-d[O2]/dt

-d[NO]/dt=rate of disappearance of NO

-d[O2]/dt=rate of disappearance of O2

Rate=[NO]^2 [O2]

putting values from the table,

1.41*10^-2 M/s=(0.0126 M)^2 (0.0125M)...(.1)

required rate=(0.0615 M)^2 (0.0175M).......(2)

dividing equation (1) by eqn (2)

1.41*10^-2 M/s/required rate=(0.0126 M)^2 (0.0125M)/(0.0615 M)^2 (0.0175M)

solving,

required rate=4.7*10^-1 M/s

Rate=[NO]^2 [O2]=-1/2 d[NO]/dt

or,Rate=-1/2 d[NO]/dt=4.7*10^-1 M/s

           - d[NO]/dt=2*4.7*10^-1 M/s=9.4*10^-1 M/s=rate of disappearance of NO

Rate=-d[O2]/dt=4.7*10^-1 M/s=rate of disappearance of O2

part E)

f=exp(-Ea/RT)

ln f=-Ea/RT

let

ln f (high T)=-Ea/RT2

ln f (low T)=-Ea/RT1

ln f (high T)-ln f (low T)=-Ea/RT2-(-Ea/RT1)

ln f (high T)/(low T)=-Ea/R(1/T2-1/T1)

ln f (high T)/(low T)=-175 *1000 j/mol/8.314 J/k mol (1/535K-1/515K)

                           =-21048.83300 K(0.00186-0.00194)K-1

                             =-21048.83300*(-0.00008)

           ln f (high T)/(low T) =1.6839

              f (high T)/(low T) =exp(1.6839)=5.3865

           

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