CHEMISTRY MIDDLE SCHOOL

In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be the concentration of A (in M) at t = 99.2 s? Only enter the numerical value with three significant figures in the answer box below. Do NOT type in the unit (M).

Answers

Answer 1

Answer:

The value of the concentration of A is 0.529.

Since the rate constant has units of s⁻¹, we can tell that the order of the reaction is 1.

Rate of first -order reaction,

$(A)=A_(0) e^(-kt)$

Here, $A_(0)$ is the initial concentration, t is the time, A is the final concentration, and k is the rate constant.

Substitute the given value values in the equation as follows,

$(A)=0.548M e^{-3.6*10^{-4s^(-1)*99.2s } } \n=0.529M$

The concentration of A in (M) is 0.529.

Find more information about first- order reaction here,

brainly.com/question/13309369

Answer 2

Answer:

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the rate constant has units of s⁻¹, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^(-kt)$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M* e^{-3.6\cdot 10^(-4)/s*99.2s}$

$[A]=0.529M$

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Answers

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