The position of equilibrium lies to the right | vedclass

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(A) In any acid-base equilibrium,the equilibrium lies towards the side of the weaker acid and weaker base.$1$. For $N_2H_5^+ + NH_3 \rightleftharpoons

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$HBr > N_2H_5^+ > NH_4^+$

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(A) In any acid-base equilibrium,the equilibrium lies towards the side of the weaker acid and weaker base.$1$. For $N_2H_5^+ + NH_3 \rightleftharpoons NH_4^+ + N_2H_4$,since the equilibrium lies to the right,$N_2H_5^+$ is a stronger acid than $NH_4^+$.$2$. For $N_2H_4 + HBr \rightleftharpoons N_2H_5^+ + Br^{-}$,since the equilibrium lies to the right,$HBr$ is a stronger acid than $N_2H_5^+$.Combining these two observations,we get the order of acidic strength as $HBr > N_2H_5^+ > NH_4^+$.

The position of equilibrium lies to the right in each of these reactions:
$N_2H_5^+ + NH_3 \rightleftharpoons NH_4^+ + N_2H_4$
$NH_3 + HBr \rightleftharpoons NH_4^+ + Br^{-}$
$N_2H_4 + HBr \rightleftharpoons N_2H_5^+ + Br^{-}$
Based on this information,what is the order of acidic strength?

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The position of equilibrium lies to the right in each of these reactions:$N_2H_5^+ + NH_3 \rightleftharpoons NH_4^+ + N_2H_4$$NH_3 + HBr \rightleftharpoons NH_4^+ + Br^{-}$$N_2H_4 + HBr \rightleftharpoons N_2H_5^+ + Br^{-}$Based on this information,what is the order of acidic strength?