For the following gaseous reversible reaction:
\(3\text{A}_{(g)} + \text{B}_{(g)} \rightleftharpoons \text{A}_3\text{B}_{(g)} \quad (\Delta_r H = -q\text{ kJ})\). The amount of product \(\text{A}_3\text{B}_{(g)}\) is affected by ____

The given reversible reaction is \(3\text{A}_{(g)} + \text{B}_{(g)} \rightleftharpoons \text{A}_3\text{B}_{(g)}\) with \(\Delta_r H = -q\text{ kJ}\).

The reaction is exothermic since \(\Delta_r H < 0\). According to Le Chatelier's principle, a change in temperature will shift the equilibrium position, thereby affecting the amount of product \(\text{A}_3\text{B}_{(g)}\).

The change in the number of gaseous moles is \(\Delta n_g = 1 - (3 + 1) = -3\). Since \(\Delta n_g \neq 0\), a change in pressure will also shift the equilibrium position, affecting the amount of product.

A catalyst only alters the rate at which equilibrium is attained and does not change the equilibrium composition or the amount of product formed.

Therefore, the amount of product \(\text{A}_3\text{B}_{(g)}\) is affected by both temperature and pressure.

Answer: Both temperature and pressure