Question

2h2o2(g)→2h2o(g)+o2(g) what will decrease the equilibrium amount of H2O

Answer 1

$2H_2O_2(g)\to$ $2H_2O(g)+O_2(g)$

The equilibrium amount of $[H_2O]$ will decrease when the reaction proceed in backward direction or we can say that when the concentration of $H_2O_2$ is increased on reactant side then only the concentration of product side will decrease only . if we write the $k_{c}$ of above reaction then we can write down as $-$

$k_c=\dfrac{[H_2O]^{2}[O_2]}{[H_2O_2]^{2}}$

we can also explain by le chatelier's principle where the concentration or mole reaction is more reaction proceed in that direction only if we increase the concentration of $[H_2O_2]$ then obviously we will see decrease in concentration of $[H_2O].$