Question

The decomposition reaction of carbon disulfide to carbon monosulfide and sulfur is first order with k = 2.80 ✕ ✕ 10−7 sec-1 at 1000°c. cs2(g) → cs(g) + s(g)
a. how much of a 4.83-gram sample of carbon disulfide would remain after 37.0 days? 1.97 1.97 grams carbon disulfide
b. how much carbon monosulfide would be formed after 37.0 days? 1.14 1.65 grams carbon monosulfide useful information 1.013 bar = 760 torr = 1 atm = 760 mm hg

Answer 1

Answer: a) 1.97 grams of carbon disulfide will remain after 37.0 days.

              b) 2.85 grams of carbon monosulfide will be formed after 37.0 days.

Explanation: The decomposition of carbon disulfide is given as:

                          $CS_2(g)\rightarrow CS(g)+S(g)$

at t=0                    4.83g             0          0

at t=37 days        4.83 - x            x           x

here,

x = amount of $CS_2$ utilised in the reaction

This reaction follows first order kinetics so the rate law equation is:

$k=\frac{2.303}{t}log\frac{A_o}{A}$

where, k = rate constant

t = time

$A_o$ = Initial mass of reactant

A = Final mass of reactant

a) For this, the value of

$k=2.80\times10^{-7}sec^{-1}$

t = 370 days = 3196800 sec

$A_o$ = 4.83

A = 4.83-x

Putting values in the above equation, we get

$2.8\times 10^{-7}sec^{-1}=\frac{2.303}{3196800sec}log\left(\frac{4.83}{4.83-x}\right)$

x = 2.85g

Amount of $CS_2$ remained after 37 days = 4.83 - x

                                                                     = 1.97g

b) Amount of carbon monosulfide formed will be equal to "x" only which we have calculated in the previous part.

Amount of carbon monosulfide formed = 2.85g