# (Answered): Please answer but also explain all of them! Thank you! Consider the reaction: 4NH_3 + 7O_…

Please answer but also explain all of them! Thank you!

Consider the reaction: 4NH_3 + 7O_2 rightarrow 4NO_2 + 6H_2O 1. At a certain instant the initial rate of disappearance of the oxygen gas is X. What is the value of the appearance of water at the same instant? A) 1.2 X B) 1.1 X C)0.86 X D) 0.58 X E) cannot be determined from the data For a reaction in which A and B react to form C, the following initial rate data were obtained: What is the rate law? A) Rate = k[A][B] B) Rate = k[A]^2[B] C) Rate = k[A][B]^2 D) Rate = k [A]^2[B]^2 E) Rate = k[A]^3 The reaction 2NO rightarrow N_2 + O_2 has the following rate law. – Delta [NO]/Delta t = 2k[NO]^2 After a period of 2.7 times 10^3 s, the concentration of NO falls from an initial value of 2.8 times 10^-3 mol/L to 2.0 times 10^- 3 mol/L. What is the rate constant, K? A) 1.5 times 10^-7 M^-1 middot s^-1 B) 5.3 times 10^-2 M^-1 middot s^-1 C) 3.2 times 10^-1 middot s^-1 D) 1.2 times 10^-4 M^-1 middot s^-1 E) 2.6 times 10^-2 M^-1 middot S^-1

## Expert Answer

Answer

1.for the reaction 4NH3+7O2———>4NO2+6H2O

-d(O2/dt)/7= d(H2O)/6dt

where 7 and 6 are coefficients of O2 and H2O

d(H2O)/dt= -(6/7) *(dO2)/dt= (6/7)*X = 0.86X ( C is the correct answer)

2.

Le the rate law be represented as

-rA= K[A] a [B]b, where a and b are orders of reaction with respect to A and B, K is the rate constant.

From the 1st data point,

1= K[0.1]a [0.1]b   (1)

From the 2nd data point,

4= K[0.1]a [0.2]b   (2)

Eq.2/Eq.1 gives 4= 2b, b=2

From the 3rd data point

8= K[0.2]a [0.2]b   (2)

Eq.3/Eq.2 , 2= 2a, a=1

Substituting the value of a and b in Eq.1

So the rate equation is –rA= K[A] [B]2 ( C is the correct answer)

3.

Given –d[NO]/dt= 2K[NO]2

-d[NO]/[[NO]2= 2K*dt

When integrated

1/[NO]= 2Kt+C, where C is integration constant

At t=0 ,[NO]=[NO]o, intital concentration of NO

1/[NO]0 =C

Hecne

1/[NO]= 1/[NO]o+ 2Kt (1)

Given t=2.7*1000sec, [NO]o= 2.8*10-3 Mol/L, [NO]= 2*10-3 Mol/L

Substituting these values in Eq.1

1/(2*10-3)= 1/(2.8*10-3)+ 2*2700K

K= 0.026455/M.sec ( E is the correct answer)

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