The first-order rate constant for the decomposition of N2O5 to NO2 and O2 at 70 °C is 5.3 × 10–3 s–1. Suppose we start with 0.500 mol of N2O5(g) in a 0.700 L container, how many moles of N2O5 will remain after 1.5 min?
ln(A0At)=ktln(\frac{A_0}{A_t})=ktln(AtA0)=kt
ln(A0At)=5.3×10−3×1.5×60ln(\frac{A_0}{A_t})=5.3×10^{-3}×1.5×60ln(AtA0)=5.3×10−3×1.5×60
ln(A0At)=0.447ln(\frac{A_0}{A_t})=0.447ln(AtA0)=0.447
At=(0.500e0.447)=0.31molesA_t=(\frac{0.500}{e^{0.447}})=0.31 molesAt=(e0.4470.500)=0.31moles
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments