JEE MAIN - Chemistry (2021 - 27th August Morning Shift - No. 14)
The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is
$$2{K_2}C{r_2}{O_7} + 8{H_2}S{O_4} + 3{C_2}{H_6}O \to 2C{r_2}{(S{O_4})_3} + 3{C_2}{H_4}{O_2} + 2{K_2}S{O_4} + 11{H_2}O$$
If the rate of appearance of Cr2(SO4)3 is 2.67 mol min$$-$$1 at a particular time, the rate of disappearance of C2H6O at the same time is _____________ mol min$$-$$1. (Nearest integer)
$$2{K_2}C{r_2}{O_7} + 8{H_2}S{O_4} + 3{C_2}{H_6}O \to 2C{r_2}{(S{O_4})_3} + 3{C_2}{H_4}{O_2} + 2{K_2}S{O_4} + 11{H_2}O$$
If the rate of appearance of Cr2(SO4)3 is 2.67 mol min$$-$$1 at a particular time, the rate of disappearance of C2H6O at the same time is _____________ mol min$$-$$1. (Nearest integer)
Answer
4
Explanation
$$\left( {{{Rate\,of\,disappearance\,of\,{C_2}{H_6}O} \over 3}} \right)$$
$$ = \left( {{{Rate\,of\,disappearance\,of\,C{r_2}{{(S{O_4})}_3}} \over 2}} \right)$$
$$ \Rightarrow \left( {{{2.67\,mol/\min \, \times 3} \over 2}} \right) = $$ rate of disappearance of C2H6O.
$$\Rightarrow$$ Rate of disappearance of C2H6O = 4.005 mol/min.
$$ = \left( {{{Rate\,of\,disappearance\,of\,C{r_2}{{(S{O_4})}_3}} \over 2}} \right)$$
$$ \Rightarrow \left( {{{2.67\,mol/\min \, \times 3} \over 2}} \right) = $$ rate of disappearance of C2H6O.
$$\Rightarrow$$ Rate of disappearance of C2H6O = 4.005 mol/min.
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