- A$44\; g\ CO_2$
- B$48\; g\ O_3$
- ✓$8\; g\ H_2$
- D$64\; g\ SO_2$
$64 g$ $SO_2=1$ mole
$44 \mathrm{g} \mathrm{CO}_2=1 \mathrm{mole}$
$48 \mathrm{g} O_2=48 / 32=1.5 \mathrm{mole}$
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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)
