For the reaction 2A+3B+(3/2)C→3P,which statement is correct?

2A+3B+(3/2)C→3P ROR=12[−d[nA]dt]=13[−d[nB]dt]=23[−d[nC]dt]=12[+d[ne]dt] =[−dnAdt]=23[−d[nB]dt]=43[−d[ne]dt]

For the reaction 2A+3B+(3/2)C→3P,which statement is correct?

2A+3B+(3/2)C→3P ROR=12[−d[nA]dt]=13[−d[nB]dt]=23[−d[nC]dt]=12[+d[ne]dt] =[−dnAdt]=23[−d[nB]dt]=43[−d[ne]dt]

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For the reaction $2 A + 3 B + (3 / 2) C \to 3 P$ ,which statement is correct?

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$d n_{A} / d t = d n_{B} / d t = d n_{C} / d t$

$d n_{A} / d t = (3 / 2) d n_{B} / d t = (3 / 4) d n_{C} / d t$

$d n_{A} / d t = (2 / 3) d n_{B} / d t = (4 / 3) d n_{C} / d t$

$d n_{A} / d t = (2 / 3) d n_{B} / d t = (3 / 4) d n_{C} / d t$

answer is C.

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Detailed Solution

$2 A + 3 B + (3 / 2) C \to 3 P R O R = \frac{1}{2} [\frac{- d [n_{A}]}{d t}] = \frac{1}{3} [\frac{- d [n_{B}]}{d t}] = \frac{2}{3} [\frac{- d [n_{C}]}{d t}] = \frac{1}{2} [\frac{+ d [n_{e}]}{d t}] = [\frac{- d n_{A}}{d t}] = \frac{2}{3} [\frac{- d [n_{B}]}{d t}] = \frac{4}{3} [\frac{- d [n_{e}]}{d t}]$