In the reaction NH4NO2(aq)→ N2(g)+2H2O(l) the volume of N2 after 20 min and after a long time is 40 ml and 70 ml respectively. The value of rate constant is :

Question

In the reaction NH4NO2(aq)→ N2(g)+2H2O(l) the volume of N2 after 20 min and after a long time is 40 ml and 70 ml respectively. The value of rate constant is : (A) (1/20)In(7/4) min-1 (B) (2.303/1200)log(7/3) sec-1 (C) (1/20)log(7/3) min-1 (D) (2.303/20)log(11/7) min-1

Tardigrade Admin · Accepted Answer

NH4NO2(aq)→ N2(g)+2H2O(l) K=(2.303/t)log((V∞-VO/V∞-Vt)) ⇒ V0=0 ⇒ K=(2.303/20)log ((70-0/70-40)) =(2.303/20)log((70/30)) ⇒ K=(2.303/1200)log(7/3)

Q. In the reaction $N H_{4} N O_{2} (a q) \to N_{2} (g) + 2 H_{2} O (l)$ the volume of $N_{2}$ after 20 min and after a long time is $40$ ml and $70$ ml respectively. The value of rate constant is :

$(1/20) I n (7/4) mi n^{- 1}$

$(2.303/1200) l o g (7/3) se c^{- 1}$

$(1/20) l o g (7/3) mi n^{- 1}$

$(2.303/20) l o g (11/7) mi n^{- 1}$

Q. In the reaction NH4​NO2​(aq)→N2​(g)+2H2​O(l) the volume of N2​ after 20 min and after a long time is 40 ml and 70 ml respectively. The value of rate constant is :

(1/20)In(7/4)min−1

(2.303/1200)log(7/3)sec−1

(1/20)log(7/3)min−1

(2.303/20)log(11/7)min−1

NH4​NO2​(aq)→N2​(g)+2H2​O(l) K=t2.303​log(V∞​−Vt​V∞​−VO​​) ⇒V0​=0⇒K=202.303​log(70−4070−0​) =202.303​log(3070​) ⇒K=12002.303​log37​

Q. In the reaction $N H_{4} N O_{2} (a q) \to N_{2} (g) + 2 H_{2} O (l)$ the volume of $N_{2}$ after 20 min and after a long time is $40$ ml and $70$ ml respectively. The value of rate constant is :

$(1/20) I n (7/4) mi n^{- 1}$

$(2.303/1200) l o g (7/3) se c^{- 1}$

$(1/20) l o g (7/3) mi n^{- 1}$

$(2.303/20) l o g (11/7) mi n^{- 1}$