For the reaction below, would you expect there to be a difference in the equilib

Question

For the reaction below, would you expect there to be a difference in the equilibrium position if the reaction was done in a closed container vs in a container open to the atmosphere? Why or why not? Think about partial pressures and maintaining equilibrium. Mg(s) + H_2O(l) MgO(s) + H_2(g) + heat For the reactions below, list two ways you could shift the equilibrium position towards the right (favoring the formation of products). heat + 2NOBr(g) 2NO(g) + Br_2(g) 2CO(g) + O_2(g) 2CO_2(g) + heat CO_0(g) + H_2O(1) H_2CO_3(aq)

Explanation / Answer

8 A)

a) Heat is added , so it is endothermic reaction.

Hence,

On increasing temperature , equilibrium shifts to right.

b) On addition of reactant NOBr, equilibrium shifts to right.

B)

a) Heat is released , so it is exothermic reaction.

Hence,

On decreasing temperature , equilibrium shifts to right.

b) On addition of reactants CO or O2 , equilibrium shifts to right.

C)

a) On addition of reactants CO2 or H2O , equilibrium shifts to right.

b) On removal of product H2CO3 , equilibrium shifts to right.

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