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The reaction shifts to the left (towards reactants). What happens to the amount of [tex]$H_2O$[/tex] as the reaction shifts?

[tex]\[2 NO (g) + H_2(g) \rightleftharpoons N_2O (g) + H_2O (g) + \text{heat}\][/tex]

A. The amount of [tex]$H_2O$[/tex] goes down.
B. The amount of [tex]$H_2O$[/tex] does not change.
C. The amount of [tex][tex]$H_2O$[/tex][/tex] goes up.


Sagot :

To determine what happens to the amount of [tex]\( \mathrm{H_2O} \)[/tex] when the reaction shifts to the left, we need to analyze the given chemical equilibrium:

[tex]\[ 2 \mathrm{NO(g)} + \mathrm{H_2(g)} \rightleftharpoons \mathrm{N_2O(g)} + \mathrm{H_2O(g)} + \text{heat} \][/tex]

1. Understand the shift direction:
- When a reaction shifts to the left, it means that the equilibrium is moving towards the reactants.

2. Identify the components involved:
- Reactants: [tex]\( 2 \mathrm{NO(g)} \)[/tex] and [tex]\( \mathrm{H_2(g)} \)[/tex]
- Products: [tex]\( \mathrm{N_2O(g)} \)[/tex] and [tex]\( \mathrm{H_2O(g)} \)[/tex]
- Heat is produced in the forward direction, which implies it is endothermic in the reverse direction.

3. Effect on quantities of substances:
- When the equilibrium shifts to the left, the reaction will produce more reactants ([tex]\( \mathrm{NO(g)} \)[/tex] and [tex]\( \mathrm{H_2(g)} \)[/tex]) and consume the products ([tex]\( \mathrm{N_2O(g)} \)[/tex] and [tex]\( \mathrm{H_2O(g)} \)[/tex]).

4. Specifically for [tex]\( \mathrm{H_2O(g)} \)[/tex]:
- Since shifting the reaction to the left implies moving towards the reactants and consuming more of the products, the amount of [tex]\( \mathrm{H_2O(g)} \)[/tex] will decrease as the reaction shifts to the left.

Thus, the correct answer is:
A. The amount of [tex]\( \mathrm{H_2O} \)[/tex] goes down.