Let's analyze the given reaction:
[tex]\[ 2 NH_{3(s)} \leftrightarrow N_{2(s)} + 3 H_{2(s)} \][/tex]
When considering the effect of pressure changes in a system involving gases, we use Le Chatelier's Principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.
Here, the reaction involves solid ammonia ([tex]\(NH_3\)[/tex]) converting to solid nitrogen ([tex]\(N_2\)[/tex]) and three moles of solid hydrogen ([tex]\(H_2\)[/tex]). However, the main consideration should be the number of moles of gas on each side:
- Reactants: 2 moles of solid [tex]\(NH_3\)[/tex]
- Products: 1 mole of solid [tex]\(N_2\)[/tex] + 3 moles of solid [tex]\(H_2\)[/tex] = 4 moles of gas
Following Le Chatelier's Principle, increasing the pressure in the system shifts the equilibrium position to the side with fewer moles of gas to decrease the pressure. In this reaction:
- The reactants' side has 2 moles of gas.
- The products' side has 4 moles of gas.
Given that the forward reaction turns 2 moles of [tex]\(NH_3\)[/tex] into 4 moles of gas, increasing the pressure will cause the equilibrium to shift towards the reactants [tex]\(NH_3\)[/tex]. As a result, the forward reaction will decrease because the system aims to reduce the number of gas molecules by favoring the reactants' side.
Hence, the effect of increasing pressure on the forward reaction will be:
The reaction rate decreases.
