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.
