- A$\ce{BeCO_3}$
- B$\ce{MgCO_3}$
- C$\ce{CaCO_3}$
- ✓$\ce{BaCO_3}$
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$2 Cu ( s )+ H _2 O ( g ) \longrightarrow Cu _2 O ( s )+ H _2( g )$
$P _{ H _2}$ is the minimum partial pressure of $H _2$ (in bar) needed to prevent the oxidation at $1250 K$. The value of $\ln \left( p _{ H _2}\right)$ is. . . . .
(Given: total pressure $=1$ bar, $R$ (universal gas constant) $=8 JK ^{-1} mol ^{-1}, \ln (10)=2.3$. $Cu ( s )$ and $Cu _2 O ( s )$ are mutually immiscible.
At $1250 K : 2 Cu ( s )+1 / 2 O _2( g ) \longrightarrow Cu _2 O ( s ) ; \Delta G ^\theta=-78,000 J mol ^{-1}$
$H _2( g )+1 / 2 O _2( g ) \longrightarrow H _2 O ( g ) ; \Delta G ^\theta=-1,78,000 J mol ^{-1} ; G$ is the Gibbs energy)
$(i)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - {C^ \mathbf{-} }} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(ii)$ $H _{2} C = CH - CH _{2}$
$(iii)$ $HC \equiv \stackrel{\ominus}{ C }$
$(iv)$ $\stackrel{\ominus}{ CH _{3}}$
$(v)$ $\stackrel{\ominus}{{ }_{ CN }}$

Reason : Addition of $HBr$ on $2-$ butene follows Markovnikov rule.