Questions: For the reaction CH3OH(aq) + H+(aq) + Cl-(aq) -> CH3Cl(aq) + H2O(l) the value of [H+]was measured over a period of time. Given the data in the table, find the average rate of disappearance of H+(aq) for the time interval between each measurement. Interval: 0 s to 31.0 s t(s) [H+](M) 0 2.38 31.0 2.16 66.0 2.04 120.0 1.87 reaction rate: M / s Interval: 31.0 s to 66.0 s reaction rate: M / s Interval: 66.0 s to 120.0 s reaction rate: M / s

Solution Steps

The reaction involves the disappearance of \(\mathrm{H}^{+}\) ions over time. We are given the concentration of \(\mathrm{H}^{+}\) at different time intervals and need to calculate the average rate of disappearance for each interval.

The average rate of disappearance of a reactant is given by the formula: \[ \text{Rate} = -\frac{\Delta [\mathrm{H}^{+}]}{\Delta t} \] where \(\Delta [\mathrm{H}^{+}]\) is the change in concentration of \(\mathrm{H}^{+}\) and \(\Delta t\) is the change in time.

• \(\Delta [\mathrm{H}^{+}] = 2.16 - 2.38 = -0.22 \, \text{M}\)
• \(\Delta t = 31.0 - 0 = 31.0 \, \text{s}\)
• Rate = \(-\frac{-0.22}{31.0} = 0.0070968 \, \text{M/s}\)

• \(\Delta [\mathrm{H}^{+}] = 2.04 - 2.16 = -0.12 \, \text{M}\)
• \(\Delta t = 66.0 - 31.0 = 35.0 \, \text{s}\)
• Rate = \(-\frac{-0.12}{35.0} = 0.0034286 \, \text{M/s}\)

• \(\Delta [\mathrm{H}^{+}] = 1.87 - 2.04 = -0.17 \, \text{M}\)
• \(\Delta t = 120.0 - 66.0 = 54.0 \, \text{s}\)
• Rate = \(-\frac{-0.17}{54.0} = 0.0031481 \, \text{M/s}\)

Final Answer