Half lite and rate constant Use data from Figure to evaluate the (a) half-life a
ID: 550387 • Letter: H
Question
Half lite and rate constant Use data from Figure to evaluate the (a) half-life and (b) rate constant for the first-order decomposition of N2Os at 67°C: N2Os(g)2 NO2(g)202(g) (a) When the molarity of N2Os drops to one-half its initial value, also drops to one-half its initial value because partial pressure and molarity are proportional. As the pressure drops from 800 mmHg to 400 mmHg, one-half the N20s originally present is consumed. This occurs in about 120 s, thus t12 120 s. The pressure drops to 200 mmHg at about 240 s: 2 x t/2 240 s. Again, t120s, confirming that the decomposition of N2Os is first order 500 200 (b) We determine the rate constant k using t= 120 s in rate Equation 0.693 0.693 120 s 5.8 × 10-3 s-1 57 1/2Explanation / Answer
a) The rate law of the reaction is
ln(Ao/At) = kt
We know the value of k is equal to 0.693/120 and At = 1/16 Ao, susbtituting the values we get
ln(Ao/(Ao/16)) = 0.693/120 * t
ln(16) = 0.693/120 * t
t = 4 * 120 = 480 seconds
b)
10 min = 10 * 60 = 600 seconds
ln(Ao/At) = kt
ln(Ao/At) = 0.693/120 * 600
ln(Ao/At) = 5 * 0.693
At = Ao/32 = 0.15g
Hence N2O5 left after 10 minutes is equal to 0.15g
c)
Concentration left after 6 minutes will be
ln(Ao/At) = 0.693/120 * 6 * 60
At = Ao/8 = 4.80/8 = 0.60 grams
Hence the pressure of N2O5 will be dropped to 800/2^3 = 100 mm Hg
Pressure of NO2 will be 700* 2 = 1400 mm Hg
Pressure of O2 will be 700/2 = 350 mm Hg
Total Pressure = 1850 mm Hg
For N2O5 alone, 100 mm Hg
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