$H_{3} P O_{4}_{1} H^{+} + H_{2} P O_{4}^{-}; K_{1} H_{2} P O_{4}^{-}_{2} H^{+} + H P O_{4}^{2 -}; K_{2} H P O_{4}^{2 -}_{3} H^{+} + P O_{4}^{3 -}; K_{3}$

The equilibrium constant for, $H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}$ will be:

Q. For a Ist order decomposition of A as given

Therefore rate constant (K) for the overall decomposition of A is

Q. For two reactions, activation energies are

E_{a_{1}}

and

E_{a_{2}}

; rate constants are

k_{1}

and

k_{2}

at the same temperature. If

k_{1} > k_{2}

then

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The activation energies of two reaction are Ea and E′a in the order Ea>E′a. If the temperature of the reacting systems is increased from T1 to T2, predict which alternative is correct K' is rate constant at a higher temperature. Assume A being same for both the reactions:

The correct option is A K1<K2 and K′1<K′2As we know,More is energy of activation, lesser is rate constant(K=Ae−Ea/RT)Thus, since Ea>E′aK1<K2 and K′1<K′2.

The correct option is A $K_{1} < K_{2}$ and $K_{1}^{'} < K_{2}^{'}$
As we know,
More is energy of activation, lesser is rate constant
$(K = A e^{- E_{a} / R T})$
Thus, since $E_{a} > E_{a}^{'}$
$K_{1} < K_{2}$ and $K_{1}^{'} < K_{2}^{'}$ .