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Juliette [100K]
2 years ago
7

What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if

Chemistry
2 answers:
Sav [38]2 years ago
5 0

Answer:

C. The rate would be one-fourth.

Another option would be:

The rate would be four times larger

Depending on what class you're taking.

Explanation:

Ede4ka [16]2 years ago
3 0

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.  

Can be formulated:  

Reaction: aA ---> bB  

\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}

or  

\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}

The concentration of NO were halved, so the rate :

\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1

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What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a
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144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm  [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K  [At STP]

Now first find the number of moles

F₂  +  CaBr₂  →  CaF₂  +  Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So,  199.89 g CaBr₂ reacts with  = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

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