Question

Carbon disulfide, a poisonous, flammable liquid, is an excellent solvent for phosphorus, sulfur, and some other nonmetals. A kinetic study of its gaseous decomposition reveals these data: Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s) 1 0.100 2.7 × 10−7 2 0.080 2.2 × 10−7 3 0.055 1.5 × 10−7 4 0.044 1.2 × 10−7 (a) Choose the rate law for the decomposition of CS2

Answer 1

Answer:

Rate law: = $k[CS_2]^1$

Explanation:

Given:

t               $[CS_2]$

0.100            $2.7 \times 10^{-7}$

0.080          $2.2 \times 10^{-7}$

0.055         $1.5\times 10^{-7}$

0.044         $1.2\times 10^{-7}$

Rate law for the given reaction: $k[CS_2]^n$

Where, n is the order of the reaction.

Divide rate 1 with rate 3

$\frac{0.100}{0.055} =\frac{k[CS_2 (1)]^n}{k[CS_2 (3)]^n} \\\frac{0.100}{0.055} =\frac{k[2.7 \times 10^{-7}]^n}{k[1.5\times 10^{-7}]^n}\\1.81=[1.8]^n\\ n=1$

So, rate law = $k[CS_2]^1$

Answer 2

Answer:

$r=-3.73x10^{5}s^{-1} [CS_2]$

Explanation:

Hello,

In this case, a linealization helps to choose the rate law for the decomposition of CS₂ as it is generalized via:

$r=-k[CS_2]^{n}$

Whereas $n$ accounts for the order of reaction, which could be computed by linealizing the given data using the following procedure:

$-ln(r)=ln(k[CS_2]^{n})\\-ln(r)=ln(k)+ln([CS_2]^{n})\\-ln(r)=ln(k)+n*ln([CS_2])$

Therefore, on the attached picture you will find the graph and the lineal equation wherein the slope is the order of the reaction and the y-axis intercept the natural logarithm of the rate constant. In such a way, the order of reaction is 1 and the rate constant is:

$ln(k)=12.83\\k=exp(12.83)\\k=3.73x10^{5}s^{-1}$

Best regards.