Question

Photobromination of
cinnamic acid to
dibromocinnamic acid was
carried out in blue light of
wavelength 440 nm at
35°C using light
a
intensity of 1.5 X 10-3 J
per second. An exposure of
20 minutes produced
decrease of 0.075 millimole
of bromine. The solution
absorbed 80 percent of the
light passing through it.
Calculate the quantum yield
of the reaction.​

Answer 1

The quantum yield of the reaction is 1420.

Given:

Photobromination of cinnamic acid to

dibromocinnamic acid was carried out in blue light of wavelength 440 nm at 35°C using light intensity of 1.5 X 10-3 joule per second. An exposure of 20 minutes produced a decrease of 0.075 millimoles

of bromine. The solution absorbed 80 percent of the light passing through it.

To Find:

The quantum yield of the reaction.

Solution:

To find the quantum yield we will follow the following steps:

As we know,

The energy of a 1 mole photon is calculated by the formula:

$\frac{0.12}{wavelength \: in \: metre} = \frac{0.12}{440 \times {10}^{ - 9} } \: j {mol}^{ - 1}$

Also,

1 min = 60 seconds

So,

20 min = 20 × 60 = 1200sec

Intensity =

$\frac{energy}{time}$

Energy in 20 minutes or 1200 seconds will be

$energy = intensity \times time = 1.5 \times {10}^{ - 3} \times 1200 = 1.8 \: joule$

Also, 80% of this energy is used in dissociation. so,

The energy of dissociation =

$\frac{80}{100} \times 1.8 = 1.44 \: joule$

Energy is given by 1-mole photon

$= \frac{0.12}{440 \times {10}^{ - 9} } \: j {mol}^{ - 1}$

So,

The number of moles should be produced by 1.44 joules of energy =

$\frac{1}{\frac{0.12}{440 \times {10}^{ - 9} } \: j {mol}^{ - 1} } \times 1.44 = \frac{1.44 \times 440 \times {10}^{ - 9} }{0.12 } = 0.528 \times {10}^{ - 5} \: moles$

An exposure of 20 minutes produced a decrease of 0.075 millimoles of bromine which means 0.075 moles are produced.

Now,

Quantum yield =

$\frac{number \: of \: moles \: produced}{total \: number \: of \: moles \: should \: produce \:} = \frac{0.075 \times {10}^{ - 3} }{0.528 \times {10}^{ - 5} } = 14.20 \times {10}^{2} = 1420$

Henceforth, the quantum yield of the reaction. is 1420.

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Answer 2

Answer:

The quantum yield of the given reaction = $1420$

Explanation:

Given :

$Cinnamic$ $acid$ is converted into $dibromocinnamic\ acid$ through photobromination.

The reaction was carried out in blue light of wavelength $440nm$ at a temperature of $30$°$c$.

The intensity of light = $1.5$×$10^{-3} J$.

When exposed for $20\min$ there is decrease of $0.075\ millimole$.

Solution has absorbed $80\%$ of light.

To find : The quantum yield of the reaction.​

Solution :

The energy of one mole of photon

= $0.12/wavelength(m)$ = $0.12/440$ × $10^{-9} J \ mol ^{-1}$

$20\ min = 1200\ sec$

Intensity = $Energy/Time$

$Energy = Intensity$ × $Time$

$Energy$ = $1.5$ × $10^{-3}$ × $1200 = 1.8\ J$

As $80\%$ of energy is used in the case of dissociation, then energy of dissociation becomes $1.44 J$

So the number of moles created by $1.44 J$ of energy =

$1/2.7$ × $10^{5}$$j\ mol^{-1}$ × $1.44$ = $0.528$ × $10^{-5} \ moles$

so in reaction exposure of $20 minutes$ created a decrease of $0.075$ $millimoles$ of bromine that indirectly means $0.075 \ moles$ are created.

Quantum Yield = $no\ of\ moles\ produced / no\ of\ moles\ should\ be\ produced=0.075/0.528$ × $10^{2}$

                                                                                        $=14.2$ × $10^{2}$

                                                                                        $=1420$

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