Chemistry, 16.01.2020 07:31 tommyaberman

The gas-phase decomposition of ch3cho (g) occurs according to the equation
ch3cho (g) > ch4 (g) +co (g) and is second order. the value of the rate constant is 0.105 m-1 x s-1 at 490 degrees celcius. if the concentration of ch3cho (g) is 0.012 m initially, what will be its concentration 5.0 minutes later.

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Answer from: 20009

Assuming that temperature and the rate law stayed unchanged, the concentration after 5.0 minutes would be $\rm 8.7\times 10^{-3}\; mol \cdot L^{-1}$ .

Explanation:

The rate law is a function that takes the concentrations of the reactants as its input and return the rate of the reaction at that given time. In this case, since there's only one reactant, the rate law would be in the form:

$\text{Rate} = k \cdot [{\rm CH_3CHO\; (g)}]^{n}$ ,

where

$k = \rm 0.105\; mol^{-1}\cdot L\cdot s^{-1}$ is the rate constant, and

is the order of the reactant $[{\rm CH_3CHO\; (g)}]$ in this reaction.

The question stated that this reaction is of second order. In other words, the order of this reaction is . By definition, the order of a reaction is the sum of the orders of all the reactants in the rate law. In this case, since $[\rm CH_3CHO\; (g)]$ is the only reactant, its should must also be . That is: n = 2 and $\text{Rate} = k \cdot [{\rm CH_3CHO\; (g)}]^2$ .

This rate law can be solved as a differential equation. To keep the expressions simple, let (with the unit $\rm mol \cdot L^{-1}$ ) represent the concentration of $[\rm CH_3CHO\; (g)]$ . That is: let $x = [\rm CH_3CHO\; (g)]$ .

The reaction rate of a chemical reaction is the opposite of first derivative of concentration with respect to time (in seconds.) (There should be a negative sign, since the concentration of the reactants was supposed to decrease.) In this case,

$\displaystyle \text{Rate} = -\frac{d}{dt}\, \left([\rm CH_3CHO\; (g)]\right) = -\frac{dx}{dt}$ .

That would be the left-hand side of the rate law. The right-hand side would become:

$k\cdot [{\rm CH_3CHO\; (g)}]^2 = k \, x^2$ .

Equate the two halves to obtain the differential equation:

$\displaystyle -\frac{dx}{dt} = k\, x^2$ .

Group the two variables with their respective differentials. That is: rewrite the equation such that is on the same side as while is on the same side as . Note that in this case there's no such variable as . Simply leave alone on one side of the equation and all the and on the other side.

$\displaystyle -\frac{1}{x^2}\, dx = k\, dt$ .

Integrate both sides (indefinitely, i.e., without knowing what the bounds are:)

$\displaystyle \int\left[-\frac{1}{x^2}\, dx\right] = \int\left[k\, dt\right]$ .

For the left-hand side, apply the power rule for integration. For the right-hand side, keep in mind that is simply a constant and is not dependent on the value of . Since this integration is indefinite, add a constant of integration to one side of the equation.

$\displaystyle \frac{1}{x} = k\, t + C$ .

Rearrange the equation to obtain an expression of concentration at time :

$\displaystyle x = \frac{1}{k\, t + C}$ .

Substitute in the value for :

$\displaystyle x = \frac{1}{0.105\, t + C}$ .

At time (initial condition,) the concentration is $\rm 0.012\;mol \cdot L^{-1}$ . Hence, the constant here should ensure that when t = 0 , x = 0.012 . Solve the equation for :

$\displaystyle 0.012 = \frac{1}{C}$ .

$\displaystyle C = \frac{1}{0.012}$ .

$\displaystyle x = \frac{1}{0.105\, t + \dfrac{1}{0.012}}$ .

Evaluate this expression at $t = 5.0\times 60 = 300\; \rm s$ .

$\displaystyle x = \frac{1}{0.105\times 300 + \dfrac{1}{0.012}} \approx 8.7\times 10^{-3}\; \rm mol \cdot L^{-1}$ .

That's the concentration five minutes after the initial condition.

Answer from: Quest

did anyone actually do this? like if anyone has submitted this i rly need .

Answer from: Quest

The answer to this problem is. 67

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