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The given chemical equation is:

On balancing the equation we get,

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:
Δ![H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]](https://tex.z-dn.net/?f=H_%7Breaction%7D%5E%7B0%7D%3C%2Fp%3E%3Cp%3E%3D%5BH_%7Bf%7D%5E%7B0%7D%28Al_%7B2%7DO_%7B3%7D%28s%29%29%20%2B%20%283%2AH_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DSO_%7B4%7D%28aq%29%29%5D%20-%20%20%20%5BH_%7Bf%7D%5E%7B0%7D%28Al_%7B2%7DSO_%7B4%7D%28aq%29%29%20%2B%20%283%2AH_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%28l%29%29%5D)
=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]
=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]
=-98.21kJ/mol
Total enthalpy change when 15 mol of
reacts will be=

Answer:
7.5 sec.
Explanation:
A = A₀e⁻^kt => k = ln(A/A₀)/-t => ln(6.25/100)/-30s = 0.0924 s⁻¹
k·t₀.₅=0.693 => t₀.₅=0.693/k = (0.693/0.0924)s = 7.5 s