Respuesta :
The first order reaction for decomposition of carbon disulfide is as follows:
[tex]CS_{2}(g)\rightarrow CS(g)+S(g)[/tex]
The rate constant for the above reaction is [tex]2.80\times 10^{-7}sec^{-1}[/tex].
(a) The expression for rate constant for first order reaction is as follows:
[tex]k=\frac{2.303}{t}log\frac{A_{0}}{A_{t}}[/tex]
Here, k is rate constant, t is time of the reaction, [tex][A_{0}][/tex] is initial concentration of reactant and [tex][A_{t}][/tex] 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, [tex]37 days=37\times 86400 sec=319680 sec[/tex]
Now, putting the values in expression for rate constant,
[tex]2.80\times 10^{-7}sec^{-1}=\frac{2.303}{319680 sec}log\frac{4.83 g}{mt}[/tex]
On rearranging,
[tex]0.38866=log\frac{4.83 g}{m_{t}}[/tex]
Taking antilog both sides,
[tex]2.4472=\frac{4.83 g}{m_{t}}[/tex]
Or,
[tex]m_{t}=\frac{4.83 g}{2.4472}=1.97 g[/tex]
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 [tex]CS_{2}[/tex] 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:
[tex]m_{reacted}=m_{0}-m_{t}=(4.83-1.97)g=2.86 g[/tex]
Calculating number of moles from it,
[tex]n=\frac{m}{M}[/tex]
Molar mass of [tex]CS_{2}[/tex] is 76.14 g/mol
Putting the values,
[tex]n=\frac{2.86 g}{76.14 g/mol}=0.0375 mol[/tex]
Since, 1 mol of [tex]CS_{2}[/tex] 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:
[tex]m=n\times M=0.03756 mol\times 44.07 g/mol=1.65 g[/tex]
Therefore, mass of carbon monosulfide formed after 37 days is 1.65 g.
