Question

For the following reaction

\[\ce{2X +Y ->[K] P}\]

The rate of reaction is \[\ce{\frac{d[P]}{dt} = K[X]}\]. Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement (s) about the reaction is (are)

(Use In 2 = 0.693).

विकल्प

• The rate constant K of the reaction is 13.86 × 10⁻⁴ s⁻¹.

• Half-life of ‘X’ is 50 s.

• At 50 s, \[\ce{- \frac{d[X]}{dt}}\] = 13.86 × 10⁻³ mol L⁻¹ s⁻¹.

• At 100 s, \[\ce{- \frac{d[Y]}{dt}}\] = 3.46 × 10⁻³ mol L⁻¹ s⁻¹.

Answer 1

Half-life of ‘X’ is 50 s, At 50 s, \[\ce{- \frac{d[X]}{dt}}\] = 13.86 × 10⁻³ mol L⁻¹ s⁻¹ and At 100 s, \[\ce{- \frac{d[Y]}{dt}}\] = 3.46 × 10⁻³ mol L⁻¹ s⁻¹.

Explanation:

Rate = \[\ce{\frac{d[P]}{dt} = K[X]}\]

\[\ce{- \frac{d[X]}{dt}}\] = K₁[X] = 2K[X] = 2K = K₁

\[\ce{- \frac{d[Y]}{dt}}\] = K₂[X] = K[X] = K = K₁

2K = \[\ce{\frac{1}{50} ln 2}\]

K = \[\ce{\frac{1}{100} ln 2}\]

= \[\ce{\frac{0.693}{100}}\]

= 6.93 × 10⁻³ s⁻¹

(t_1/2)_X = \[\ce{\frac{ln 2}{K_1}}\]

= \[\ce{\frac{ln 2 \times 100}{2 \times 0.693}}\]

=  50 sec

At 50 s

\[\ce{- \frac{d[X]}{dt}}\] = 2K[X]

= \[\ce{2 \times \frac{0.693}{100} \times 1}\]

= 13.86 × 10⁻³ mol L⁻¹ s⁻¹

At 100 s,

\[\ce{- \frac{d[Y]}{dt}}\] = K₂[X] = K[X]

= \[\ce{\frac{0.693}{100} \times \frac{1}{2}}\]    ...(∵ Concentration of X after two half-lives = \[\ce{\frac{1}{2}}\] M )

= 3.46 10⁻³ mol L⁻¹ s⁻¹