The equilibrium reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} + [\text{H}^+] + \text{OH}^- \rightleftharpoons 2\text{CrO}_4^{2-}. \]
In a basic medium (OH$^-$), the reaction shifts to the right to produce more CrO$_4^{2-}$, as OH$^-$ consumes H$^+$, reducing its concentration. Conversely, in an acidic medium (H$^+$), the equilibrium shifts to the left, favoring Cr$_2$O$_7^{2-}$.
Thus, the equilibrium shifts to the right in a basic medium.
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.