a) Explain the following terms: i) Rate of reaction ii) Activation energy of a reaction b) The decomposition of phosphine, PH$_3$, proceeds according to the equation $4\mathrm{PH_3(g)}\rightarrow\mathrm{P_4(g)}+6\mathrm{H_2(g)}$. It is found that the reaction follows the rate equation: Rate = k[PH$_3$]. The half-life of PH$_3$ is 37.9 s at 120°C. i) How much time is required for 3/4th of the PH$_3$ to decompose? ii) What fraction of the original sample of PH$_3$ remains behind after 1 minute?

CBSE Class 12 · 2010 · All India · Chemical KineticsLong Answer

a) Explain the following terms:
i) Rate of reaction
ii) Activation energy of a reaction
b) The decomposition of phosphine, PH

_{3}

, proceeds according to the equation

4 P H_{3} (g) \to P_{4} (g) + 6 H_{2} (g)

. It is found that the reaction follows the rate equation: Rate = k[PH

_{3}

]. The half-life of PH

_{3}

is 37.9 s at 120°C.
i) How much time is required for 3/4th of the PH

_{3}

to decompose?
ii) What fraction of the original sample of PH

_{3}

remains behind after 1 minute?

Answer: (a) Defined below. (b)(i) ≈ 75.8 s (ii) ≈ 1/3 of the original sample.

Solution

(a)(i) Rate of reaction: the change in concentration of a reactant or product per unit time, e.g. rate

= - \frac{Δ [ R ]}{Δ t}

.
(ii) Activation energy: the minimum extra energy over the average energy that reactant molecules must acquire to form the activated complex and change into products.
(b) Rate

= k [P H_{3}]

— first order:

k = \frac{0.693}{t _{1/2}} = \frac{0.693}{37.9} = 0.0183 s^{- 1}

(i) For 3/4 decomposition,

[R] = [R]_{0} /4

t = \frac{2.303}{0.0183} lo g 4 = \frac{2.303 \times 0.6021}{0.0183}

t \approx 75.8 s (= 2 t_{1/2})

(ii) After

t = 60

lo g \frac{[ R ] _{0}}{[ R ]} = \frac{0.0183 \times 60}{2.303} = 0.4768 \Rightarrow \frac{[ R ] _{0}}{[ R ]} \approx 3

\frac{[ R ]}{[ R ] _{0}} \approx \frac{1}{3}

CBSE Class 12 Board · 2010 · All India · Subjective

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CBSE Class 12 · 2010 · All India · Chemical KineticsLong Answer

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