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Rate constant and remaining BOD

Short answer

From y5/y11 ratio, solving (1-10^(-11k))/(1-10^(-5k))=1.3 yields k25=0.1118/day (base10). Ultimate BOD L=345.4 mg/L. At30°C, k30=0.1406/day; remaining organic matter after6 days = L·10^(-k30·6) =49.5 mg/L.

A wastewater sample was incubated at 25°C. 5 day BOD of sample was equal to 250 mg/l and 11 day BOD of sample was found to be 325 mg/l. Calculate the rate reaction constant and organic matter remaining in sample after 6 days at 30°C.

Given at 25°C: y5=250y_5=250 mg/L, y11=325y_{11}=325 mg/L. Assume a first-order reaction in base 10, yt=L(1−10−kt)y_t=L(1-10^{-kt}), and θ=1.047\theta=1.047.

Step 1: Rate constant at 25°C

Divide the two equations to remove LL:

y11y5=1−10−11k1−10−5k=325250=1.3\frac{y_{11}}{y_5}=\frac{1-10^{-11k}}{1-10^{-5k}}=\frac{325}{250}=1.3

Solving by trial: k25=0.1k_{25}=0.1 gives 1.346, k25=0.2k_{25}=0.2 gives 1.104; refining gives

k25=0.1118 day−1 (base 10)k_{25}=0.1118\ \text{day}^{-1}\ (\text{base }10)

Step 2: Ultimate BOD

L=2501−10−0.1118×5=345.4 mg/LL=\frac{250}{1-10^{-0.1118\times5}}=345.4\ \text{mg/L}

Step 3: At 30°C

k30=k25(1.047)5=0.1118×1.2582=0.1406 day−1k_{30}=k_{25}(1.047)^{5}=0.1118\times1.2582=0.1406\ \text{day}^{-1}

Organic matter remaining after 6 days (the ultimate BOD is the same at any temperature):

L6=L⋅10−k30×6=345.4×10−0.8439=49.5 mg/LL_6=L\cdot10^{-k_{30}\times6}=345.4\times10^{-0.8439}=49.5\ \text{mg/L}

Answer: k25k_{25} = 0.1118 day⁻¹; organic matter remaining after 6 days at 30°C = 49.5 mg/L (BOD exerted = 295.9 mg/L).

✓ Answer checked by Claude Opus 5.5 (high effort)

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