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For the system at equilibrium, which of the following are correct

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On increasing the temperature of an endothermic reaction, the concentration in moles per litre of the reactants increase

$\log K = \frac{1}{2.303 R} (Δ S - \frac{Δ H}{T})$

On increasing the temperature of an endothermic reaction, the equilibrium shifts in forward direction because $K$ increases

On increasing the temperature of an endothermic reaction, the equilibrium shifts in forward direction because $Q$ decreases

answer is A, C.

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Detailed Solution

$\begin{array}{l} Δ G = Δ H - T Δ S = - 2.303 R T \log K \\ T Δ S - Δ H = 2.303 R T \log K \Rightarrow \end{array}$

$\frac{T Δ S - Δ H}{2.303 R T} = \log K \Rightarrow \frac{1}{2.303 R} [Δ S - \frac{Δ H}{T}] = \log K$

Hence choice .(a) is correct.

On changing temperature K changes and not Q

Also $\log \frac{K_{2}}{K_{1}} = \frac{Δ H}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

For endothermic reactions on increasing T, K increases the concentration of products.

Thus choice (c) is correct while (b) and (d) are not correct

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