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Chemistry Question on Chemical Kinetics

For the following reaction 2X+Y→kP2 X+Y \xrightarrow{k} P the rate of reaction is d[P]dt=k[X]\frac{d[P]}{d t}=k[X] Two moles of XX are mixed with one mole of YY to make 10 L10\, L of solution At 50 s50 \,s, 0505 mole of YY is left in the reaction mixture The correct statement(s) about the reaction is(are) (Use : ln⁡ 2=0693)\ln\,2 = 0693)

A

The rate constant, kk, of the reaction is 13.86×10−4s−113.86 \times 10^{-4} s ^{-1}.

B

Half-life of XX is 50 s50\, s.

C

At 50 s,−d[X]dt=13.86×10−3mol L−1s−150\, s,-\frac{d[X]}{d t}=13.86 \times 10^{-3} mol\, L^{-1} s^{-1}.

D

At 100 s,−d[Y]dt=3.46×10−3mol L−1s−1100 \,s,-\frac{d[Y]}{d t}=3.46 \times 10^{-3} mol\, L^{-1} s^{-1}.

Answer

At 100 s,−d[Y]dt=3.46×10−3mol L−1s−1100 \,s,-\frac{d[Y]}{d t}=3.46 \times 10^{-3} mol\, L^{-1} s^{-1}.

Explanation

Solution

D

At 100 s,−d[Y]dt=3.46×10−3mol L−1s−1100 \,s,-\frac{d[Y]}{d t}=3.46 \times 10^{-3} mol\, L^{-1} s^{-1}.