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Standard XII

Chemistry

Arrhenius Acid

Radiolysis of...

Question

Radiolysis of water yields hydrogen according to the reaction:
$2 H_{2} O ⟶ H_{2} + H_{2} O_{2}$
Calculate the value of hydrogen produced per hour in a 200 kW reactor. The yield of hydrogen is 1.7 molecules per 100 eV of energy absorbed.

2846.5 L at STP

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4546.5 L at STP

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3246.5 L at STP

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None of these

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Solution

The correct option is B 2846.5 L at STP
Power of reactor=200kW
Energy absorbed in 1 second=200kJ
Energy absorbed in 1 hour $= 200 \times 3, 600 = 7.2 \times 10^{5} k J$

$= \frac{7.2 \times 10^{5} \times 1000}{1.6 \times 10^{- 19}} e V = 4.5 \times 10^{27} e V$
No. of molecules of hydrogen produced $= \frac{4.5 \times 10^{27} \times 1.7}{100} = 7.65 \times 10^{25}$
Moles of $H_{2}$ gas produced $\frac{7.65 \times 10^{25}}{6.023 \times 10^{23}} = 127.0131 m o l e s$

Volume released at STP (using PV=nRT)

$= \frac{127.0131 \times 0.0821 \times 273}{1} L$

$= 2846.7 L ≅ 2846.5 L$

Thus correct option is A.

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Similar questions

Radiolysis of water yields

H_{2}

as given below:

2 H_{2} O ⟶ H_{2} O_{2} + H_{2}

In the analysis, it is found that

1.7

molecules of

H_{2}

are produced per

100

eV of energy absorbed. A nuclear power reactor of

200

kW capacity has been installed based on this reaction.

What mass of heavy water is required to yield

448

kilolitres of heavy hydrogen at STP?

Q. Calculate the volume of

H_{2}

gas at STP which would be produced per minute from radiolysis of water in a 1000 kW reactor, assumimg that 100 eV of energy produces 1 molecule of hydrogen

2 H_{2} O ⟶ H_{2} + H_{2} O_{2}

(1 e V = 1.6 \times 10^{- 19} J)

Plan energy given in kW is to be converted into

J s^{- 1}

and thus, volume of

H_{2}

gas produced in this reactor can be calculated.

In a fuel cell, hydrogen and oxygen react to produce electricity. In the process, hydrogen gas is oxidized at anode and oxygen at cathode. If 67.2 L of $H_{2}$ at STP reacts in 15 min, the average current (in A) produced is :

Anode reaction : $H_{2} + 2 ⊖ O H \to 2 H_{2} O + 2 e^{-}$

Cathode reaction :

O_{2} + 2 H_{2} O + 2 e^{-} \to 4 ⊖ O H

(write your answer to nearest integer)

Q. Anodic oxidation of ammonium hydrogen sulphate produces ammonium persulphate.

N H_{4} H S O_{4} \to N H_{4} S O_{4}^{-} + H^{+}

2 N H_{4} S O_{4}^{-} \to (N H_{4})_{2} S_{2} O_{8} + 2 e^{-}

(Anodic oxidation)

2 H^{+} + 2 e^{-} \to H_{2}

(Cathodic reduction)
Hydrolysis of ammonium persulphate forms

H_{2} O_{2}

(N H_{4})_{2} S_{2} O_{8} + 2 H_{2} O \to 2 N H_{4} H S O_{4} + H_{2} O_{2}

Current efficiency inn electrolytic process is

60

%. Calculate the amount of current required to produce

85 g

H_{2} O_{2}

per hour.
Hydrolysis reaction shows

100

% yield.

Q. Assertion :Radiolysis of water yields

H_{2}

. Reason: The reaction during radiolysis of water is disproportionation reaction.

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Radiolysis of water yields hydrogen according to the reaction:2H2O⟶H2+H2O2Calculate the value of hydrogen produced per hour in a 200 kW reactor. The yield of hydrogen is 1.7 molecules per 100 eV of energy absorbed.

Radiolysis of water yields hydrogen according to the reaction:
$2 H_{2} O ⟶ H_{2} + H_{2} O_{2}$
Calculate the value of hydrogen produced per hour in a 200 kW reactor. The yield of hydrogen is 1.7 molecules per 100 eV of energy absorbed.