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ICSE Class 10 Chemistry Question 4 of 37

Mole Concept and Stoichiometry — Question 16

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Question

Question 15

Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:

2H2O ⟶ 2H2 + O2

(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?

(b) The 2500 cm3 of H2 is subjected to 2122\dfrac{1}{2} times increase in pressure (temp. remaining constant). What volume will H2 now occupy?

(c) Taking the value of H2 calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.

Answer

2H2O⟶2H2+O2 Vol.2 Vol.1 Vol.\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow & 2\text{H}_2 & + &\text{O} \\ 2\text{ Vol.} & & 2\text{ Vol.}& & 1\text{ Vol.} \end{matrix}

2 Vol. of water gives 2 Vol. of H2 and 1 Vol. of O2

∴ If 2500 cm3 of H2 is produced, volume of O2 produced = 25002\dfrac{2500}{2} = 1250 cm3

(b) V1 = 2500 cm3

P1 = 1 atm = 760 mm

T1 = T

T2 = T

P2 = [760 x 2 12\dfrac{1}{2}] + [760] = 760 [52\dfrac{5}{2} + 1] = 760 x 72\dfrac{7}{2} = 2660 mm

V2 = ?

Using formula:

P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1} = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}

760×2500T\dfrac{760 \times 2500}{\text{T}} = 2660×V2T2660 \times \dfrac{\text{V}_2}{\text{T}}

V2 = 760×25002660\dfrac{760 \times 2500 }{2660} = 50007\dfrac{5000}{7}

(c) V1 = 50007\dfrac{5000}{7} = 714.29 cm3

P1 = P2 = P

T1 = T

V2 = 2500 cm3

T2 = ?

Using formula:

P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1} = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}

P×714.29T\dfrac{\text{P} \times 714.29}{\text{T}} = P×2500T2\dfrac{\text{P} \times 2500}{\text{T}_2}

T2 = 2500714.29\dfrac{2500 }{714.29} x T

T2 = 3.5 x T

Hence, T2 = 3.5 times T or temperature should be increased by 3.5 times