Question

If equal concentrations are mixed (of products & reactants) then which of the reaction will proceed to the right and which will proceed to the left.
$HC{O}_3^{-}\left(aq\right)+S{O}_4^{2-}\rightleftharpoons HS{O}_4^{-}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$.

Answer 1

${HCO}_3^{-}\left(aq\right)+{SO}_4^{-2}\rightleftharpoons {HSO}_4^{-}\left(aq\right)+{CO}_3^{-2}\left(aq\right)$

If equal concentration mixed (both reactants and products)

$Q=\frac{\left[{HSO}_4^{-}\right]\left[{CO}_3^{-2}\right]}{\left[{HCO}_3^{-}\right]\left[{SO}_4^{-2}\right]}=1$ ($\because$ all are equal)

$K$ for the reaction is $<<1$ ($\because {HCO}_3^{(-)}$ is weak acid than ${HSO}_4^{-}$)

Amount or size of flame depends upon heat supplied, amount of fuel. Here heat of reaction is constant and at equilibrium amount of fuel is constant. Hence equilibrium is reason for constant flame size.