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3 years ago

What mass of H2SO4 is contained in 60.00 mL of a 5.85 M solution of sulfuric acid

Chemistry

1 answer:

strojnjashka [21]3 years ago

5 0

First, we have to calculate the number of moles of H2SO4 in the solution:

V=60 mL = 0.06 L

c=5.85 mol/L

n=V×c=0.06×5.85=0.351 mol

Then we need to find the molar mass of H2SO4:

2×Ar(H) + Ar(S) + 4×Ar(O) =

=2 + 32 + 64 = 98 g/mol

Finally, we need to find the mass of H2SO4:

m=0.351 × 98 = 34.398 g

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0.0907 s

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This an Arrhenius equation problem, so you relate the half-life with the kinetic constant of the reaction in order to calcule the same thermodynamic parameters at another temperature.

To calcule the kinetic constant of the reaction you need to know the order of it, look closely to the sentence "The value of the half-life is independent of the inital concentration of N2O present." the only order independent from the initial concentration of reagents is first order, so you can calculate K at 800 K, using:

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Now you can use Arrhenius equation to calcule K at 1150.66 K

$ln(\frac{k1}{k2} )=-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} )$

$k2= k1*exp(-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} ))=3.239*10^{-5}s^{-1}*exp(-\frac{270 000 J/mol }{8.314 J/mol *k }(\frac{1}{1150.66K} - \frac{1}{800K} ))=7.639 s^{-1}$

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