Question

What is the concentration (in mol L$^{-1}$) of the product after 20 s in the following reaction. Given that A $\rightarrow$ 3 B , rate = k[A]$^0$ Time(s) \(\quad\) Concentration of the reactant (mol L$^{-1}$) 0 \(\quad\) \(\quad \quad \quad 0.1 15 \quad \quad \quad \quad 0.05 20 \quad \quad \quad \quad 0.1-x\)

• A. $6.6 \times 10^{-2}$
• B. $1.32 \times 10^{-3}$
• C. $1.98 \times 10^{-1}$
• D. $2.2 \times 10^{-2}$

Hint:

Identify the order of reaction from the rate law (rate = k[A]$^0$ is zero order).
For zero-order reaction: $[A]_t = [A]_0 - kt$. Rate constant $k = ([A]_0 - [A]_t)/t$.
Calculate $k$ using the given concentration data at two time points.
Calculate the amount of A reacted ($x$) at the desired time ($t=20$s): $x = kt$.
Use stoichiometry to find concentration of product B. If A $\rightarrow$ 3B, then $[B]_{formed} = 3x$.
Be precise with fractions if possible to avoid rounding errors until the final step. $0.05/15 = 1/300$.

Answer 1

The Correct Option is C

The reaction is A $\rightarrow$ 3B. The rate law is given as rate = k[A]$^0$. This means it is a zero-order reaction with respect to A. For a zero-order reaction, the rate is constant and independent of the concentration of A: rate = $k$. The integrated rate law for a zero-order reaction is: $[A]_t = [A]_0 - kt$ where $[A]_t$ is the concentration of A at time $t$, $[A]_0$ is the initial concentration of A, and $k$ is the rate constant. From the given data: At $t=0 \text{ s}$, $[A]_0 = 0.1 \text{ mol L}^{-1}$. At $t=15 \text{ s}$, $[A]_{15} = 0.05 \text{ mol L}^{-1}$. We can use this information to find the rate constant $k$. $0.05 = 0.1 - k(15)$. $15k = 0.1 - 0.05 = 0.05$. $k = \frac{0.05}{15} \text{ mol L}^{-1}\text{s}^{-1}$. $k = \frac{5 \times 10^{-2}}{15} = \frac{1}{3} \times 10^{-2} \text{ mol L}^{-1}\text{s}^{-1} \approx 0.00333 \text{ mol L}^{-1}\text{s}^{-1}$. Now, we need to find the concentration of A at $t=20 \text{ s}$, which is given as $0.1-x$. This $x$ is the amount of A reacted. $[A]_{20} = [A]_0 - k(20)$. $[A]_{20} = 0.1 - \left(\frac{0.05}{15}\right)(20)$. $[A]_{20} = 0.1 - \frac{0.05 \times 20}{15} = 0.1 - \frac{1}{15} = 0.1 - \frac{100}{15 \times 100} = 0.1 - \frac{1}{15}$. $\frac{1}{15} \approx 0.0666...$ $[A]_{20} = 0.1 - 0.0666... = 0.0333... \text{ mol L}^{-1}$. So, $0.1-x = 0.0333...$, which means $x = 0.1 - 0.0333... = 0.0666... \text{ mol L}^{-1}$. This $x$ is the decrease in concentration of A, which is the amount of A reacted. The reaction is A $\rightarrow$ 3B. For every 1 mole of A that reacts, 3 moles of B are formed. So, the concentration of product B formed, $[B]_{formed}$, is $3 \times (\text{amount of A reacted})$. Amount of A reacted by time $t=20\text{s}$ is $x = [A]_0 - [A]_{20} = kt$. $x = (\frac{0.05}{15})(20) = \frac{1}{15} \text{ mol L}^{-1} \approx 0.0666... \text{ mol L}^{-1}$. Concentration of B formed at $t=20 \text{ s}$: $[B]_{20} = 3x = 3 \times \frac{1}{15} = \frac{3}{15} = \frac{1}{5} \text{ mol L}^{-1}$. $[B]_{20} = 0.2 \text{ mol L}^{-1}$. Let's check the options: % Option (a) $6.6 \times 10^{-2} = 0.066$. This is $x$. % Option (b) $1.32 \times 10^{-3} = 0.00132$. % Option (c) $1.98 \times 10^{-1} = 0.198$. This is very close to $0.2$. % Option (d) $2.2 \times 10^{-2} = 0.022$. The calculated concentration of product B is $0.2 \text{ mol L}^{-1}$. Option (c) is $0.198 \text{ mol L}^{-1}$. The slight difference might be due to rounding or if "0.1-x" implies $x$ itself is the answer for something. "What is the concentration (in mol L$^{-1}$) of the product after 20 s". This means $[B]_{20}$. My value is $0.2$. Option (c) is $0.198$. It is likely the intended answer through some rounding of $k$ or the values. $k = 1/300 \approx 0.003333...$ Amount of A reacted $x = kt = (1/300) \times 20 = 20/300 = 2/30 = 1/15$. Conc. of B $= 3x = 3 \times (1/15) = 1/5 = 0.2$. Let's check if $x$ in $0.1-x$ refers to something else. The table gives "Concentration of the reactant (mol L$^{-1}$)" At $t=20$, it is $0.1-x$. This $x$ is not defined further in the question context. However, usually $x$ represents the amount reacted from initial. So $[A]_{20} = [A]_0 - x_{reacted}$. Here $x_{reacted} = kt = (1/15)$. So $[A]_{20} = 0.1 - (1/15)$. The problem wrote $0.1-x$ for $[A]_{20}$. This suggests $x$ is the amount of A reacted by $t=20s$. My calculation $x_{reacted} = 1/15 \approx 0.0666...$ Then $[B] = 3 \times x_{reacted} = 3 \times (1/15) = 1/5 = 0.2$. It's possible the options are constructed with more precision in $k$. $k = 0.05/15 = (5/100)/15 = 5/(1500) = 1/300$. This is exact. Amount of A reacted in 20s $= k \times 20 = (1/300) \times 20 = 20/300 = 1/15$. Amount of B formed $= 3 \times (1/15) = 3/15 = 1/5 = 0.2$. The value $0.198$ is $19.8 \times 10^{-2}$ or $1.98 \times 10^{-1}$. $0.198 = 198/1000 = 99/500$. $1/5 = 100/500$. They are close. $0.198$ is $0.2 \times 0.99$. Perhaps there is a $3.3%$ error somewhere or a slightly different $k$. If initial conc. was $0.1$, and $k=(0.1-0.05)/15 = 0.05/15$. If values were like $t=0, [A]=0.1$; $t=15, [A]=0.0505$. Then $k=(0.1-0.0505)/15 = 0.0495/15 = 0.0033$. Then $x_{reacted} = 0.0033 \times 20 = 0.066$. $[B] = 3 \times 0.066 = 0.198$. This matches option (c). This means the value $0.05$ at $t=15$s might imply some precision limit. If $k=0.0033$ (from $0.0495/15$), then $0.0495$ change in 15s. The data $0.1 \rightarrow 0.05$ in $15s$ implies $k = (0.1-0.05)/15 = 0.05/15$. This is exact. My calculation of $0.2$ is robust from the given numbers. Option (c) $0.198$ is likely what the question setter derived due to using an approximation for $1/3$ like $0.33$ or $0.0033$ for $k/10$. Rate constant $k = 0.05/15 = 1/300 \text{ mol L}^{-1}\text{s}^{-1}$. Amount of A reacted at $t=20\text{s}$ is $x = k \cdot t = (1/300) \cdot 20 = 20/300 = 1/15 \text{ mol L}^{-1}$. Concentration of B formed is $3x = 3 \cdot (1/15) = 1/5 = 0.2 \text{ mol L}^{-1}$. The closest option is (c) $0.198 \text{ mol L}^{-1}$. The discrepancy is $(0.2 - 0.198)/0.2 \times 100% = (0.002/0.2) \times 100% = 1%$. This is likely a precision/rounding issue in the question's intended solution path. \[ \boxed{0.198 \text{ mol L}^{-1} \text{ (closest option to calculated 0.2 mol L}^{-1}\text{)}} \]