Photo AI

Methanol and hydrochloric acid react according to the following balanced equation: CH₃OH(aq) + HCl(aq) → CH₃Cl(aq) + H₂O(l) 5.1 State TWO factors that can INCREASE the rate of this reaction - NSC Physical Sciences - Question 5 - 2016 - Paper 2

Question icon

Question 5

Methanol-and-hydrochloric-acid-react-according-to-the-following-balanced-equation:--CH₃OH(aq)-+-HCl(aq)-→-CH₃Cl(aq)-+-H₂O(l)--5.1-State-TWO-factors-that-can-INCREASE-the-rate-of-this-reaction-NSC Physical Sciences-Question 5-2016-Paper 2.png

Methanol and hydrochloric acid react according to the following balanced equation: CH₃OH(aq) + HCl(aq) → CH₃Cl(aq) + H₂O(l) 5.1 State TWO factors that can INCREASE... show full transcript

Worked Solution & Example Answer:Methanol and hydrochloric acid react according to the following balanced equation: CH₃OH(aq) + HCl(aq) → CH₃Cl(aq) + H₂O(l) 5.1 State TWO factors that can INCREASE the rate of this reaction - NSC Physical Sciences - Question 5 - 2016 - Paper 2

Step 1

5.1 State TWO factors that can INCREASE the rate of this reaction.

96%

114 rated

Answer

  1. Increase the temperature: Raising the temperature generally increases the kinetic energy of the particles, leading to more frequent and effective collisions between the reactants.

  2. Increase the concentration of HCl: A higher concentration of reactants increases the number of particles in a given volume, resulting in an increased likelihood of collisions.

Step 2

5.2 Define the term reaction rate.

99%

104 rated

Answer

The reaction rate is defined as the change in concentration of a reactant or product over time. It reflects how quickly reactants are converted into products in a chemical reaction.

Step 3

5.3.1 Calculate the average reaction rate, in (mol dm⁻³) min⁻¹ during the first 15 minutes.

96%

101 rated

Answer

To calculate the average reaction rate, we use the formula:

ext{Average Rate} = - rac{ ext{Change in Concentration}}{ ext{Time}}

From the data:

  • Initial concentration at 0 minutes = 1.90 mol dm⁻³
  • Final concentration at 15 minutes = 1.45 mol dm⁻³

Change in concentration = 1.45 - 1.90 = -0.45 mol dm⁻³.

Time interval = 15 minutes.

Now substituting into the formula:

ext{Average Rate} = - rac{-0.45 ext{ mol dm}^{-3}}{15 ext{ min}} = 0.03 ext{ (mol dm}^{-3} ext{ min}^{-1})

Step 4

5.3.2 Use the data in the table to draw a graph of concentration versus time on the attached GRAPH SHEET.

98%

120 rated

Answer

To graph the data, plot the time (in minutes) on the x-axis and the concentration of HCl (in mol dm⁻³) on the y-axis. The points to plot are:

  • (0, 1.90)
  • (15, 1.45)
  • (210, 0.60)

Since the graph is not a straight line, ensure that the curve reflects the decreasing concentration of HCl over time.

Step 5

5.3.3 At this point, determine the concentration of HCl(aq) at the 40th minute.

97%

117 rated

Answer

To estimate the concentration of HCl at the 40th minute, observe the trend in the graph. The graph should be used to interpolate between the data points. Given the decreasing concentration, the concentration of HCl at 40 minutes could be approximated around 1.20 mol dm⁻³ based on the curve drawn.

Step 6

5.3.4 Calculate the mass of CH₃Cl(aq) in the flask at the 215th minute.

97%

121 rated

Answer

At the 215th minute, the concentration of CH₃Cl can be approximated as follows:

Let’s say the concentration at the 215th minute worked out to be roughly 0.60 mol dm⁻³.

The volume of the solution = 60 cm³ = 0.060 dm³.

Using the formula:

extMass=extConcentrationimesextVolumeimesextMolarMass ext{Mass} = ext{Concentration} imes ext{Volume} imes ext{Molar Mass}

where Molar Mass of CH₃Cl = 50.5 g/mol:

extMass=0.60extmoldm3imes0.060extdm3imes50.5extg/mol=1.81extgrams ext{Mass} = 0.60 ext{ mol dm}^{-3} imes 0.060 ext{ dm}^{3} imes 50.5 ext{ g/mol} = 1.81 ext{ grams}.

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;