La fórmula correcta del óxido salino de estaño es: *

Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

3 years ago

Which corresponds to a pressure of 1.23 atm??

Vadim26 [7]

Atmospheric pressure, sometimes also called barometric pressure, is the pressure exerted by the weight of air in the atmosphere of Earth (or that of another planet)

1 atm is equivalent to = 101325 Pa

= 760 mmHg

= 760 torr

= 1.01325 bar

So 1.23 atm is equal to

= 124629.8 Pa

= 934.8 mmHg

= 934.8 torr

= 1.2462 bar

6 0

3 years ago

Does anyone know how to do number 15?

vovikov84 [41]

Answer:

yes i do 980

Explanation:

i899 1/2

6 0

3 years ago

A cat travels 240 meters in 30 seconds, and a sprinter travels 100 meters in 9.5 seconds. Which is traveling faster, the cat or

joja [24]

Answer:

The cat

Explanation:

1) You need to find how many meters per second each thing is sprinting, so you can divide the meters by seconds for each.

$\frac{240}{30} = 8 m/s\\\frac{100}{9.5} = around 10.5 m/s$

2) Per second, the sprinter is able to achieve 8 meters, while the cat can do around 10.5 meters. 10.5 is greater than 8, making the cat faster than the sprinter.

4 0

2 years ago

The standard free energy of activation of one reaction A is 95.00 kJ mol–1 (22.71 kcal mol–1). The standard free energy of activ

diamong [38]

Answer:

The answer to the questions are as follows

Reaction B is 4426.28 times faster than reaction A

(b) Reaction B is faster.

Explanation:

To solve the question we are meant to compare both reactions to see which one is faster

The values of the given activation energies are as follows

For A

Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and

for B

Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)

T is the same for both reactions and is equal to 298 k

Concentration of both reaction = 1M

The Arrhenius Law is given by

k = $Ae^{\frac{-E_{a} }{RT} }$

Where

k = rate constant

Ea = activation energy

R = universal gas constant

T = temperature (Kelvin )

A = Arrhenius factor

Therefore

For reaction A, the rate constant k₁ is given by k₁ = $Ae^{\frac{-95000}{(8.314)(298)} }$

And for B the rate constant k₂ is given by k₂ = $Ae^{\frac{-74200 }{(8.314)(298)} }$

k₁ = A×2.225×10⁻¹⁷

k₂ = A×9.850×10⁻¹⁴

As seen from the above Reaction B is faster than reaction A by (A×9.850×10⁻¹⁴)/(A×2.225×10⁻¹⁷) or 4426.28 times

3 0

3 years ago