Questions: Name: CHEM 111 Quiz 3 1) Given the reaction 2 A(g) -> B(g) + C(g) a) Express the rate of reaction in terms of the change in concentration in terms of the change in concentration of each of the reactants and products. b) When [C] is increasing at 2.0 mol / L s, how fast is [A] decreasing?

Name: 
CHEM 111 Quiz 3
1) Given the reaction 2 A(g) -> B(g) + C(g)
a) Express the rate of reaction in terms of the change in concentration in terms of the change in concentration of each of the reactants and products.
b) When [C] is increasing at 2.0 mol / L s, how fast is [A] decreasing?
Transcript text: Name: $\qquad$ CHEM 111 Quiz 3 1) Given the reaction $2 A_{(\mathrm{g})} \rightarrow B_{(\mathrm{g})}+\mathrm{C}_{(\mathrm{g})}$ a) Express the rate of reaction in terms of the change in concentration in terms of the change in concentration of each of the reactants and products. b) When [C] is increasing at $2.0 \mathrm{~mol} / \mathrm{L} \mathrm{s}$, how fast is $[A]$ decreasing?
failed

Solution

failed
failed

Solution Steps

Step 1: Express the Rate of Reaction

For the reaction \(2 A_{(\mathrm{g})} \rightarrow B_{(\mathrm{g})} + \mathrm{C}_{(\mathrm{g})}\), the rate of reaction can be expressed in terms of the change in concentration of each reactant and product.

The rate of reaction \(r\) is given by: \[ r = -\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt} = \frac{d[C]}{dt} \]

Step 2: Relate the Rate of Change of [C] to [A]

Given that \([C]\) is increasing at \(2.0 \, \mathrm{mol/L \cdot s}\), we can use the rate expression to find how fast \([A]\) is decreasing.

From the rate expression: \[ r = \frac{d[C]}{dt} = 2.0 \, \mathrm{mol/L \cdot s} \]

Since: \[ r = -\frac{1}{2} \frac{d[A]}{dt} \]

We can set the rates equal to each other: \[ 2.0 = -\frac{1}{2} \frac{d[A]}{dt} \]

Step 3: Solve for \(\frac{d[A]}{dt}\)

Solving for \(\frac{d[A]}{dt}\): \[ 2.0 = -\frac{1}{2} \frac{d[A]}{dt} \]

Multiply both sides by \(-2\): \[ \frac{d[A]}{dt} = -4.0 \, \mathrm{mol/L \cdot s} \]

Final Answer

\[ \boxed{\frac{d[A]}{dt} = -4.0 \, \mathrm{mol/L \cdot s}} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful