Should I write H_2SO_4 like a catalyst?

The nitrate ion can be identified by heating copper turnings along with concentrated sulphuric acid. Here sulfuric acid reacts with the nitrate ion to form nitric acid. Nitric acid then reacts with the copper turnings to form nitric oxide. So, does reaction equation look like this? Cu + 4 HNO_3 → Cu(NO_3)_2 + 2NO_2 +2H_2O If yes, should I write ${H}_{2} S {O}_{4}$ like a catalyst?

Mar 6, 2018

Yes.

The ${\text{H"_2"SO}}_{4}$ is present just to protonate the nitrate ions in solution, as this reaction occurs in acidic conditions.

If ${\text{H"_2"SO}}_{4}$ were to be dilute (in "catalytic amounts"), then it would not react with $\text{Cu} \left(s\right)$. If it were to be concentrated, then it would be an interfering oxidizing agent and you would get ${\text{CuSO}}_{4} \left(a q\right)$ and ${\text{SO}}_{2} \left(g\right)$ as products instead of "Cu"("NO"_3)_2(aq).

$\text{Cu"(s) + 4"HNO"_3(l) stackrel("H"_2"SO"_4" ")(->) "Cu"("NO"_3)_2(aq) + 2"NO"_2(g) + 2"H"_2"O} \left(l\right)$

When we balance this from half-reactions we might have written:

${\text{Cu"(s) -> "Cu}}^{2 +} \left(a q\right) + \cancel{2 {e}^{-}}$
$\underline{2 \left(\cancel{{e}^{-}} + 2 \text{H"^(+)(aq) + "NO"_3^(-)(aq) -> "NO"_2(g) + "H"_2"O} \left(l\right)\right)}$
$\text{Cu"(s) + 4"H"^(+)(aq) + 2"NO"_3^(-)(aq) -> "Cu"^(2+)(aq) + 2"NO"_2(g) + 2"H"_2"O} \left(l\right)$

or

$\underline{\text{Cu"(s) + 4"H"^(+)(aq) + 4"NO"_3^(-)(aq) -> "Cu"^(2+)(aq) + 2"NO"_3^(-)(aq) + 2"NO"_2(g) + 2"H"_2"O} \left(l\right)}$
$\text{Cu"(s) + 4"HNO"_3(l) -> "Cu"("NO"_3)_2(aq) + 2"NO"_2(g) + 2"H"_2"O} \left(l\right)$

Here we write ${\text{HNO}}_{3} \left(l\right)$ to indicate that it is concentrated, maybe $\text{11 M}$ or so.