What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

$2{\mathrm{K}}_2{\mathrm{CrO}}_4+2{\mathrm{HNO}}_3\to {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2{\mathrm{KNO}}_3+{\mathrm{H}}_2\mathrm{O}$The oxidation state of chromium in ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ ​ ​is +6, whereas the oxidation state of nitrogen in ${\mathrm{HNO}}_3$is +5.

Since the maximum oxidation state of nitrogen is +5, so it won’t get oxidize further. Rather it will react to form dichromate ion and a water molecule $( {\mathrm{Cr}}_2{{\mathrm{O}}_7^2}^{-} \mathrm{and} {\mathrm{H}}_2\mathrm{O} ).$

The equation involved is:

$2{\mathrm{K}}_2{\mathrm{CrO}}_4+2{\mathrm{HNO}}_3\to {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2{\mathrm{KNO}}_3+{\mathrm{H}}_2\mathrm{O}$