Answer:
The rate of the reaction is approximately 7.315974 × 10⁻⁵ mol/s
Explanation:
The volume occupied by one mole of gas at 60 °C is given from the universal gas equation as follows;
[tex]V = \dfrac{n \times R \times T}{P}[/tex]
Where;
V = Th volume occupied by the gas
n = The number of moles = 1
R = The universal gas constant = 82.05745 cm³·atm·K⁻¹·mol⁻¹
T = The temperature of the gas = 60°C = 333.15 K
P = The pressure of the gas = 1 atm
By plugging in the values, we get;
[tex]V = \dfrac{1 \ mole \times 82.05745 \ \dfrac{cm^3 \cdot atm}{K \cdot mol} \times 333.15 \ K }{1 \ atm} = 27,337.4394675 \ cm^3[/tex]
The volume of 1 mole of as at 60 °C, V ≈ 27,337.44 cm³/mole
Therefore, the number of moles, 'n', of moles of CO₂ in 30 cm³ of CO₂ is given as follows;
n ≈ 30 cm³/(27,337.44 cm³/mole) ≈ 0.0010973961 moles
Therefore;
The rate of the reaction = Δ[CO₂]/Δt ≈ 0.0010973961 moles/15 s ≈ 7.315974 × 10⁻⁵ mol/s
