Acidic medium (oxidising agent):
$[\text{Fe(CN)}_6]^{4-} + H_2O_2 + 2H^+ \to [\text{Fe(CN)}_6]^{3-} + \text{H}_2\text{O}$
H$_2$O$_2$ is reduced to H$_2$O. Other product: H$_2$O.

Alkaline medium (reducing agent):
$H_2O_2 \to H_2O + \frac{1}{2}O_2$ (H$_2$O$_2$ is oxidized)
$2[\text{Fe(CN)}_6]^{3-} + H_2O_2 + 2OH^- \to 2[\text{Fe(CN)}_6]^{4-} + O_2 + 2H_2O$
Other products: H$_2$O and O$_2$.

So the pair is: H$_2$O (acidic) and (H$_2$O + O$_2$) (alkaline).

• Hydrogen has three main isotopes:
• Protium ($^1H$): The most common, with no neutrons.
• Deuterium ($^2H$): Also known as heavy hydrogen, with one neutron.
• Tritium ($^3H$): A radioactive isotope with two neutrons.

Entropy is a state function. This means the change in entropy depends only on the initial and final states of the system, not on the path or the number of intermediate steps taken to reach that state.

Since the body is heated from an initial temperature $T_1 = 373 \text{ K}$ to a final temperature $T_2 = 473 \text{ K}$ in both cases, the entropy change $\Delta S$ of the body will be identical in both scenarios.

$\Delta S = \int \frac{dQ}{T} = \int_{T_1}^{T_2} \frac{ms dT}{T} = C \ln \left( \frac{T_2}{T_1} \right)$.

For non-zero solution, $\Delta=0$:

$\begin{vmatrix}1&k&3\\3&k&-2\\2&4&-3\end{vmatrix}=1(−3k+8)−k(−9+4)+3(12−2k)=−3k+8+5k+36−6k=44−4k=0$

So $k=11$.

With $k=11$: from equations 1 and 2: $x+11y+3z=0$ and $3x+11y-2z=0$. Subtracting: $2x-5z=0\Rightarrow x=\frac{5z}{2}$.

Substituting back: $\frac{5z}{2}+11y+3z=0\Rightarrow 11y=-\frac{11z}{2}\Rightarrow y=-\frac{z}{2}$.

$\frac{xz}{y^2}=\frac{\frac{5z}{2}\cdot z}{\frac{z^2}{4}}=\frac{\frac{5z^2}{2}}{\frac{z^2}{4}}=\frac{5}{2}\times4=\mathbf{10}$