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

The first order reaction for decomposition of carbon disulfide is as follows:

The rate constant for the above reaction is .

(a) The expression for rate constant for first order reaction is as follows:

Here, k is rate constant, t is time of the reaction, is initial concentration of reactant and is concentration at time t.

Here, concentration terms can be replaced by mass now, the mass of reactant at time 37 days should be calculated.

First convert unit of time from days to sec as follows:

1 day=86400 sec

thus,

Now, putting the values in expression for rate constant,

On rearranging,

Taking antilog both sides,

Or,

Therefore, mass of carbon disulfide remains after 37 days is 1.97 g.

(b) From the balanced chemical equation for decomposition of carbon disulfide 1 mol of gives 1 mol of CS.

now, mass of carbon disulfide remain after 37 days is 1.97 g thus, mass of carbon disulfide reacted can be calculated as follows:

Calculating number of moles from it,

Molar mass of is 76.14 g/mol

Putting the values,

Since, 1 mol of gives 1 mol of CS thus, 0.0375 mol will give 0.0375 mol of CS.

Molar mas of CS is 44.07 g/mol , calculate mass as follows:

Therefore, mass of carbon monosulfide formed after 37 days is 1.65 g.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles