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Anna007 [38]
3 years ago
8

The pressure exerted by a gas in a container depends on what?

Chemistry
1 answer:
lyudmila [28]3 years ago
7 0

Answer:

From gas laws (pressure law and Boyles law), the pressure exerted by a gas depends on Temperature of the gas and volume of the container.

Explanation:

{ \boxed{ \rm{P \:  \alpha  \:  \frac{T}{V} }}}

• P → Pressure exerted by the gas.

• T → Temperature of the gas.

• V → Volume of the container.

• from the expression, pressure exerted by the gas is directly proportional to temperature of the gas and inversely proportional to the volume of the container.

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We need to use the following formula
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E_{cell}= 0.24V
n= 4 moles
F= constant= 96500C/mol

let's plug in the values.

ΔG= -(4)(96500)(0.24)= -92640 J or -92.6 kJ
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Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]
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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}]

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