PYQ NEET- ରାସାୟନିକ ଗତିକାଳନ ଲିଙ୍କାର୍ଯ୍ୟ L-4

ପ୍ରଶ୍ନ: ଏକ ରାସାୟନିକ ପ୍ରକ୍ରିୟା ପାଇଁ#

$$ 4 \mathrm{~A}+3 \mathrm{~B} \rightarrow 6 \mathrm{C}+9 \mathrm{D} $$

$C$ ରାଇଟ୍ ଫର୍ମେସନ୍ ହେଉଛି $6 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ଏବଂ A ର ଦୂରତା ହେଉଛି $4 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$। ପ୍ରକ୍ରିୟାର ଦ୍ରୁତତା ଏବଂ 10 ସେକେଣ୍ଡର ମଧ୍ୟରେ $B$ ଶର୍ଯ୍ୟା ଶର୍ଯ୍ୟା ହେଉଛି:

A) $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ଏବଂ $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$

B) $1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ଏବଂ $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$

C) $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ଏବଂ $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$

D) $1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ଏବଂ $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$

ଉତ୍ତର: $1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ ଏବଂ $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$#