Decomposition of nitrogen pentoxide is known | vedclass

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(D) Since the given reaction is a first-order reaction,its half-life $(t_{1/2})$ remains constant.For a first-order reaction,$75\%$ decomposition occu

Vedclass · Accepted Answer

About $3\%$

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(D) Since the given reaction is a first-order reaction,its half-life $(t_{1/2})$ remains constant.For a first-order reaction,$75\%$ decomposition occurs in two half-lives $(2 	imes t_{1/2})$.Given that $75\%$ decomposition takes $24 \ minutes$,we have $2 	imes t_{1/2} = 24 \ minutes$,so $t_{1/2} = 12 \ minutes$.An hour is $60 \ minutes$,which corresponds to $60 / 12 = 5$ half-lives.The amount of reactant remaining after $n$ half-lives is given by $(1/2)^n 	imes 100\%$.For $n = 5$,the remaining amount is $(1/2)^5 	imes 100\% = (1/32) 	imes 100\% = 3.125\%$.Time $(minutes)$Remaining Amount $(\%)$$0$$100$$12$$50$$24$$25$$36$$12.5$$48$$6.25$$60$$3.125$Thus,the amount of oxide left is approximately $3\%$.Hence,Option $D$ is the correct answer.

Decomposition of nitrogen pentoxide is known to be a first order reaction. $75\%$ of the oxide had decomposed in the first $24 \ minutes$. At the end of an hour,after the start of the reaction,the amount of oxide left will be

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