For the formation of 1 mol of nitrosyl chloride at a given temperature, Delta H

Question

For the formation of 1 mol of nitrosyl chloride at a given temperature, Delta H = -35 kj. NO(g) + 1/2 CI_2(G) rightarrow NOCI(g) The activation energy for this reaction is 76 KJ/mol. What is the activation energy for the reverse reaction? A) -35 KJ/mol B) 76 KJ/mol C) -111 KJ/mol D) 111 Kj/mol E) 41 Kj/mol A 4.00-mol sample of HI is placed in a 1.00 L vessel at 460 degree C and the reaction system is allowed to come to equilibrium. The HI partially decomposes, forming 0.305 mol H_2 and 0.305 mol I_2 at equilibrium. What is the equilibrium constant K_for the following reaction at 460 Degree C? 1/2 H_2(g) + 1/2 I_2(g) HI(g) A) 8.10 times 10^-3 B) 2.75 times 10^-2 C) 6.03 D) 1.23 times 10^2 E) 11.1 For a first-order reaction, what are the possible units of the rate constant? A) L^-1 S^-1 B) S^-1 C) mol L^-1 S^-1 D) mol L^-1 E) S Consider the following reaction: 2HF(g) H_2(g) + F_2(g) (K = 1.00 times 10^-2) Given that 1.00 mol of HF (g), 0.377 mol of H_2(g) and 0.750 mol of F_2(g) are mixed in a 5.00 L flask, determine the reaction quotient, Q A) Q = 0.0566 B) Q = 0.283 C) Q = 010707 D) Q = 213 E) None of these A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH_3(g) N_2(g) + 3H_2(g) At equilibrium, it was found that the concentration of H_2 was 0.0454 M_the concentration of N_2 was 0.0151 M and the concentration of NH_3 was 0.260 M. What is K_for this equilibrium? A) 1.58 times 10^-5 B) 2.65 times 10^-3 C) 3.97 times 10^-3 D) 2.10 times 10^-5 E) 2.91 times 10^-1 The rate constant for a first-order reaction is 1.4 times 10^-2 s^-1 at 674 K and 5.3 times 10^-2 s^-1 at 830 K. what is the activation energy? A) 17 kj/mol B) 39 Kj/mol C) 4800 Kj/mol D) 17 Kj/mol E) 40 Kj/mol A second-order reaction starts with an initial concentration of 0.100 mol/L of the reactant. If the rate constant is 2.2 times 10^-2 L/(mol middot S), what is the time required to decrease the initial concentration to 0.050 mol/L? A) 2.27 S B) 31.5 S C) 450 S D) 680 S E) 1400 S

Explanation / Answer

1)

we know that

dH = activation energy of forward reaction - activation energy of reverse reaction

so

-35 = 76 - activation energy of reverse reaction

activation energy of reverse reaction = 111 kJ/mol

so

the answer is D) 111 kJ/mol

2)

we know that

concentration = moles / volume (L)

so initially

[HI] = 4 / 1 = 4 M

now

at equilibrium

[H2] = 0.305 / 1 = 0.305 M

[I2] = 0.305 / 1 = 0.305 M

now

consider the given reaction

0.5 H2 + 0.5 I2 ---> HI

using ICE table

initial concentration of H2 , I2 , HI are 0,0 , 4

change in concentration of H2,I2 ,HI are +0.5x , +0.5x , -x

equilibrium concentration of H2,I2,HI are 0.5x ,0.5x , 4-x

given

[H2]eq = 0.305

so

0.5x = 0.305

x = 0.61

so

at equilibrium

[HI] = 4 - x = 4-0.61 = 3.39

now

Kc = [HI] / [H2]^0.5 [F2]^0.5

Kc = [3.39] / [0.305]^0.5 [0.305]^0.5

Kc = 11.11

so

the answer is E) 11.1

3)

the unit of rate constant for a first order reaction is s-1

so

the answer is B) s-1

4)

we know that

concentration = moles / volume (L)

so

[HF] = 1 / 5 = 0.2

[H2] = 0.377 /5 = 0.0754

[F2] = 0.75 / 5 = 0.15

now

consider the given reaction

2HF ---> H2 + F2

the reaction quotient is given by

Q = [H2] [F2] / [HF]^2

Q = [0.0754] [0.15] / [0.2]^2

Q = 0.28275

so

the value of reaction quotient is 0.283

so

the answer is B) Q = 0.283

5)

consider the given reaction

2NH3 ---> N2 + 3 H2

the expression for Kc is given by

Kc = [N2] [H2]^3 / [NH3]^2

given

at equilibrium

[N2] = 0.0151

[H2] = 0.0454

[NH3] = 0.26

so

using those values

we get

Kc = [0.0151] [0.0454]^3 / [0.26]^2

Kc = 2.09 x 10-5

so

the answer is D) 2.1 x 10-5

6)

we know that

according to arhenius eqaution

at two different temperatures

ln (k2/k1) = (Ea/R) ( 1/T1 - 1/T2)

using given values

we get

ln ( 5.3 x 10-2 / 1.4 x 10-2) = (Ea/8.314) (1/674 - 1/830)

Ea = 39690

so

the activation energy is 40 kJ/mol

so

the answer is E) 40 kJ/mol

7)

we know that

for 2nd order reaction

1/[A] = kt + (1/[Ao])

given

initial concentation , [Ao] = 0.1

final concentration ,[A] = 0.05

rate constant = 0.022

so

1/0.05 = ( 0.022 x t) + (1/0.1)

t = 454.5

so

the time taken is 450 s

so

the answer is C) 450 s

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