A(g) + b(g) => ab(g), the rate is 0.239 mol/l. s when the initial concentrations of both a and b are 2.00 mol/l. if the reaction is first order in a and second order in b, what is the rate when the initial concentration of a = 4.81 mol/l and that of b = 2.65 mol/l give answer to 2 decimal places.

Rate ≅ 1.01 M/s (3 sig. figs.)

Explanation:

Given A(g) + B(g) => AB(g)

Rate = k[A(g)][B(g)]²

at Rate (1) = 0.239M/s = k[2.00M][2.00M]² => k = (0.239M/s) / (2.00M)(2.00M)²

k = 0.29875 M⁻²·s⁻¹

Rate (2) =  k[A(g)][B(g)]² = (0.29875M⁻²·s⁻¹)(4.81M)(2.65M)² = 1.009124472 M/s (calc. ans.) ≅ 1.01 M/s (3 sig. figs.)

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explanation:

the answer is d. 74.02