Based on the information provided, we have a reaction between hydrogen iodide (HI) gas and hydrogen gas (H₂) to form iodine gas (I₂). The equilibrium is represented by the equation:
2HI(g) = H₂(g) + I₂(g)
The concentration values given in the table correspond to the concentrations of H₂ and HI at different times.
a) From t=0 s to t=5 s: Without the specific graph mentioned in Figure 7.10, it is difficult to provide a precise explanation of the physical situation in the container during this time period. However, based on the equilibrium reaction given, we can make some general observations. At the start (t=0 s), the concentrations of H₂ and HI may be high. As time progresses, the reaction proceeds, and the concentrations of H₂ and HI may decrease while the concentration of I₂ increases. The specific behavior will depend on the rate of the forward and reverse reactions.
b) External factor altered at t=5 s: To bring about a change in the shape of the graph at t=5 s, some external factor must have been altered. The most likely factor is the total pressure within the container. Since the reaction involves gases, changes in pressure can affect the equilibrium position. However, according to the information given, a change in pressure will not affect equilibrium in this case since the number of moles of gas is the same on both sides of the equation. Therefore, if the shape of the graph changes at t=5 s, some other external factor, such as temperature or the addition of a catalyst, must have been altered.
c) Calculation of K at t=3 s: The equilibrium constant (K) can be calculated at any given time using the concentrations of the reactants and products. However, the concentrations of H₂ and HI at t=3 s are not provided in the information given. Without the necessary data, it is not possible to calculate K at t=3 s.
Lastly, the statement "1 dm³ COCI, decomposes" seems incomplete. If you provide additional information or clarify the question, I'll be happy to assist you further.