The rate of the reaction $4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\( was studied by charging \)\mathrm{PH}_{3}(g)$ into a constant-volume reaction vessel and measuring the total pressure.

The balanced chemical equation for the given reaction is already provided: \[4\mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g) + 6\mathrm{H}_{2}(g)\]

Since the reaction takes place in a constant-volume container, we can use the ideal gas law to express the relationship between pressure, moles, and temperature: \[P = \frac{nRT}{V}\] Here, \(P\) represents the pressure, \(n\) represents the number of moles, \(R\) is the ideal gas constant, \(T\) is the temperature, and \(V\) is the volume. In a constant-volume container, the total pressure is proportional to the total number of moles of gas.

Let's analyze the rate of the reaction with respect to the concentration of reactants. At the beginning of the reaction, let the moles of PH3 be \(n_0\), the moles of P4 and H2 be zero. After some time, let's say that x moles of PH3 have reacted to form P4 and H2. Based on the balanced chemical equation, we can calculate the moles of P4 and H2 formed: PH3: Moles = \(n_0 - 4x\) P4: Moles = \(x\) H2: Moles = \(6x\) The total pressure, \(P_{total}\), in the container at this point is proportional to the total number of moles: \[P_{total} = k(n_0 - 4x + x + 6x)\] Here, \(k\) is a constant of proportionality, equal to \(\frac{RT}{V}\). \[P_{total} = k(n_0 + 3x)\] As the reaction progresses, we can measure the total pressure to determine the change in moles of PH3. This will help us in understanding the rate of the reaction.