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Work Assignment: Based on the balance equation 2x + 4y -> 3z + tQ, and the following data: Time(s) 0.00 11.0, TYY(m) 1.358 0.407. a calculate A, B, C, D, E: the rate of disappearance of Y, the rate of appearance of Z, the rate of appearance of Q, the overall rate of reaction.

Chemistry

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The given reaction is:

2X+4Y→3Z+Q2X + 4Y \rightarrow 3Z + Q

The given data for reactant Y is:

  • Initial time (t1t_1) = 0.00 s0.00 \text{ s}
  • Final time (t2t_2) = 11.0 s11.0 \text{ s}
  • Initial concentration of Y ([Y]1[Y]_1) = 1.358 M1.358 \text{ M}
  • Final concentration of Y ([Y]2[Y]_2) = 0.407 M0.407 \text{ M}

Step 1: Calculate the change in concentration of Y (Δ[Y]\Delta[Y]) and the change in time (Δt\Delta t).

Δ[Y]=[Y]2−[Y]1=0.407 M−1.358 M=−0.951 M\Delta[Y] = [Y]_2 - [Y]_1 = 0.407 \text{ M} - 1.358 \text{ M} = -0.951 \text{ M} Δt=t2−t1=11.0 s−0.00 s=11.0 s\Delta t = t_2 - t_1 = 11.0 \text{ s} - 0.00 \text{ s} = 11.0 \text{ s}

Step 2: Calculate the rate of disappearance of Y.
The rate of disappearance of Y is given by −Δ[Y]Δt-\frac{\Delta[Y]}{\Delta t}.

Rate of disappearance of Y=−−0.951 M11.0 s=0.951 M11.0 s\text{Rate of disappearance of Y} = -\frac{-0.951 \text{ M}}{11.0 \text{ s}} = \frac{0.951 \text{ M}}{11.0 \text{ s}} Rate of disappearance of Y≈0.08645 M/s\text{Rate of disappearance of Y} \approx 0.08645 \text{ M/s}

Part a· 1 step

The rate of disappearance of Y is 0.0865 M/s\boxed{\text{0.0865 M/s}}

  1. Calculate the rate of disappearance of X.

    From the stoichiometry of the reaction (2X2X and 4Y4Y), the relationship between their rates is:

    −12Δ[X]Δt=−14Δ[Y]Δt-\frac{1}{2}\frac{\Delta[X]}{\Delta t} = -\frac{1}{4}\frac{\Delta[Y]}{\Delta t} Rate of disappearance of X=−Δ[X]Δt=24(−Δ[Y]Δt)=12×(Rate of disappearance of Y)\text{Rate of disappearance of X} = -\frac{\Delta[X]}{\Delta t} = \frac{2}{4} \left(-\frac{\Delta[Y]}{\Delta t}\right) = \frac{1}{2} \times (\text{Rate of disappearance of Y}) Rate of disappearance of X=12×0.08645 M/s≈0.043225 M/s\text{Rate of disappearance of X} = \frac{1}{2} \times 0.08645 \text{ M/s} \approx 0.043225 \text{ M/s}

Part b· 1 step

The rate of disappearance of X is 0.0432 M/s\boxed{\text{0.0432 M/s}}

  1. Calculate the rate of appearance of Z.

    From the stoichiometry of the reaction (4Y4Y and 3Z3Z), the relationship between their rates is:

    13Δ[Z]Δt=−14Δ[Y]Δt\frac{1}{3}\frac{\Delta[Z]}{\Delta t} = -\frac{1}{4}\frac{\Delta[Y]}{\Delta t} Rate of appearance of Z=Δ[Z]Δt=34(−Δ[Y]Δt)=34×(Rate of disappearance of Y)\text{Rate of appearance of Z} = \frac{\Delta[Z]}{\Delta t} = \frac{3}{4} \left(-\frac{\Delta[Y]}{\Delta t}\right) = \frac{3}{4} \times (\text{Rate of disappearance of Y}) Rate of appearance of Z=34×0.08645 M/s≈0.0648375 M/s\text{Rate of appearance of Z} = \frac{3}{4} \times 0.08645 \text{ M/s} \approx 0.0648375 \text{ M/s}

Part c· 1 step

The rate of appearance of Z is 0.0648 M/s\boxed{\text{0.0648 M/s}}

  1. Calculate the rate of appearance of Q.

    From the stoichiometry of the reaction (4Y4Y and 1Q1Q), the relationship between their rates is:

    11Δ[Q]Δt=−14Δ[Y]Δt\frac{1}{1}\frac{\Delta[Q]}{\Delta t} = -\frac{1}{4}\frac{\Delta[Y]}{\Delta t} Rate of appearance of Q=Δ[Q]Δt=14(−Δ[Y]Δt)=14×(Rate of disappearance of Y)\text{Rate of appearance of Q} = \frac{\Delta[Q]}{\Delta t} = \frac{1}{4} \left(-\frac{\Delta[Y]}{\Delta t}\right) = \frac{1}{4} \times (\text{Rate of disappearance of Y}) Rate of appearance of Q=14×0.08645 M/s≈0.0216125 M/s\text{Rate of appearance of Q} = \frac{1}{4} \times 0.08645 \text{ M/s} \approx 0.0216125 \text{ M/s}

Part d· 1 step

The rate of appearance of Q is 0.0216 M/s\boxed{\text{0.0216 M/s}}

  1. Calculate the overall rate of reaction.

    The overall rate of reaction is defined as:

    Overall Rate=−12Δ[X]Δt=−14Δ[Y]Δt=13Δ[Z]Δt=11Δ[Q]Δt\text{Overall Rate} = -\frac{1}{2}\frac{\Delta[X]}{\Delta t} = -\frac{1}{4}\frac{\Delta[Y]}{\Delta t} = \frac{1}{3}\frac{\Delta[Z]}{\Delta t} = \frac{1}{1}\frac{\Delta[Q]}{\Delta t}

    Using the rate of disappearance of Y:

    Overall Rate=−14Δ[Y]Δt=14×(Rate of disappearance of Y)\text{Overall Rate} = -\frac{1}{4}\frac{\Delta[Y]}{\Delta t} = \frac{1}{4} \times (\text{Rate of disappearance of Y}) Overall Rate=14×0.08645 M/s≈0.0216125 M/s\text{Overall Rate} = \frac{1}{4} \times 0.08645 \text{ M/s} \approx 0.0216125 \text{ M/s}

Part e

The overall rate of reaction is 0.0216 M/s\boxed{\text{0.0216 M/s}}

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