1
Research Question
Effect of increasing the temperature of 1 moldm−3 of methanoic acid (10.00, 15.00, 20.00, 30.00, 40.00, and 50.00
±0.01 ) in the reaction of methanoic acid(aq) and sodium hydrogen carbonate solution(aq), on the rate constant of the
change in mass measured indirectly every 15 seconds for 180 seconds (±0.01s) and then using calculations and graphing
to determine the activation energy, rate of reaction, rate expression, rate constant, and the order of reaction?
Bees contain methanoic acid in their stingers (“Bee Stings - Acid or Alkali?”, n.d.). Since methanoic acid was an
acid, the best way to neutralize it was through a bicarbonate, which in my experiment was sodium hydrogen carbonate,
commonly known as baking soda. A neutralization reaction is when an acid and base react, forming water and salt, in my
case was sodium formate. This is due to H +¿¿ andOH −¿¿ions reacting together, which produces water. An equation that
was used to demonstrate this reaction was:
methanoic acid+ sodium hydrogen carbonate solution→sodium formate+carbon dioxide +water
The chemical formula of this reaction was:
HCOOH (aq) + ¿ NaHC O 3(aq) → HCOONa❑(aq) +CO 2(g ) + H 2 O❑(l ) ¿
In my experiment, I used sodium hydrogen carbonate solution because I wanted to investigate how increasing the
temperature of methanoic acid will affect the rate constant, k, which required two liquids. The rate constant was a
constant at a certain temperature. This was because as I increased the temperature of methanoic acid, the particles in the
acid gained more kinetic energy, which allowed more successful collisions with the carbonate solution. A successful
collision was when the particles collided at the correct orientation and had energy equal to or greater than the activation
energy. In order to find a numerical value of the rate, I found the percentage decrease of the original mass at 180 seconds
of the reaction and divided it by the time, which was 180 seconds.
Next, I found the rate expression, which told me whether the methanoic acid or sodium hydrogen carbonate
solution had an impact on the rate of reaction. In the equation below, reactant A was the concentration of methanoic acid
and reactant B was the concentration of sodium hydrogen carbonate solution. The concentrations were calculated and
then substituted into the equation later when I tried to find the rate constant for each temperature:
rate=k ¿
Figure 1 General rate expression equation for my investigation
The order of reaction, represented by a (methanoic acid) and b (sodium hydrogen carbonate solution), were the
powers to which the concentration of a reactant was raised in the rate expression. I found the orders of reaction by first
halving the concentrations of each reactant. Then, I conducted the same experiment with the new concentrations in
finding the changes in mass of the solution at 30 . After that, I calculated the rate of reaction of these experiments, and
used the same method as before of dividing the percentage decrease by the time. I then compared theses rates to the
reaction rate of the experiment with the original concentrations at 30 . If the rate didn’t change when I halved the
concentrations, this was zero order. First order, was when the concentrations and rates were directly proportional to each
other, so when I halved the concentration the rate of reaction halved from the original as well. Finally, when I halved the
concentration and the rate decreased by a quarter, this was second order. Once I found the orders of reaction, I found the
rate constant for each temperature. I substituted in the values of the concentrations, orders of reaction, and rate, to solve
for k for that temperature. With the rate constant I graphed a lnk vs. 1/T graph to determine the activation energy through
−E a
the Arrhenius equation: RT , as this provided me with a straight line graph. This equation also displayed the
k=Ae
−Ea
relationship between temperature (in Kelvin) and the rate constant. The slope of this equation being helped me
R
calculate the activation energy for this specific reaction.
The experimental data demonstrated the relationship between reaction rate and concentrations. The method I
chose was to investigate the change in mass of the reaction. This was because carbon dioxide was produced, which
caused the mass to decrease. This method was chosen because of a lab experiment done in class. I investigated a similar
reaction, just with different reactants, and I wanted to investigate further this in relation to the HL Kinetics topic.
I was attacked by bees as a child, where I got stung at least 30 times, which caused me to be rushed to the
hospital. Remembering this tragic experience, I wanted to investigate how I could deal with this from a chemistry view.
Hypothesis
If I increased the temperature of one moldm−3of methanoic acid, the rate constant should increase, as rate
, 2
constant is temperature-dependent. The formula to find the rate constant relied on the orders and reaction rate. Since the
orders and concentration stayed constant in my experiment, the rate constant value relied on the reaction rate.
Next, the rate of reaction should increase as well. This meant that the percentage difference from the initial mass
of the reaction was greater at higher temperatures. This was because temperature is a factor which affected the rate of
reaction. At higher temperatures, there was fast movement of methanoic acid particles which meant that the frequency of
collisions with the sodium hydrogen carbonate solution particles increased. This was due to the particles having higher
kinetic energy and entropy (measure of disorder) which allowed them to collide more often and with more speed. This
meant the particles collided with correct orientation and energy equal to or greater than the activation energy. In addition,
every 10 the rate of reaction should double, due to the general rule.
Finally, the activation energy will stay constant, showing temperature had no effect on this variable. At higher
temperatures, the activation energy was the same, there were more particles that had energy greater than or equal to that
activation energy for successful collisions. This was represented through this Maxwell-Boltzmann Distribution Curve:
Figure 2 Maxwell Boltzmann Distribution Curve for
temperature change (Sheetal, 1970)
Variables
Independent Variable:
I changed the temperature of the methanoic acid using a Heat Plate (±0.01 ) and Ice bath
Increments:
10.00, 15.00, 20.00, 30.00, 40.00, and 50.00 measured using a data harvest temperature probe with data logger (±0.01 )
Dependent Variable:
Rate constant k, (mol−3 dm 9 s−1)
How it will be measured:
I found the rate of reaction through the average percentage difference from the original mass at 180 seconds. Then, I
found the orders of reaction for each reactant. I substituted in these values to calculate for the rate constant.
Controlled Variables
Variables I How I controlled it? Why was it important?
Controlled
Concentration I used 1 moldm−3 Concentration was a factor that affected reaction rate. As the concentration of
of methanoic methanoic acid for each methanoic acid increased, the rate of reaction increased as well. This was also the
acid experiment. case if the concentration decreased. This was due to frequencies in collisions as at
higher concentrations, there were a greater proportion of particles, which increased
the chances of successful collisions. In specific, if I increased the concentration of
methanoic acid, there was a greater proportion of these particles compared to the
carbonate solution. When reacted with the solution, the chance in these particles
successfully colliding increased with greater numbers of methanoic acid particles.
Concentration I made 0.952 moldm−3 Concentration was a factor that affected the reaction rate. As the concentration of the
of sodium of sodium hydrogen sodium hydrogen carbonate solution increased, the reaction rate increased. This was
hydrogen carbonate for the whole the case if the concentration decreased as well. In specific, if I increased the
carbonate experiment. concentration of carbonate solution, there were a greater proportion of these particles
solution perdm 3compared to methanoic acid. When reacted with the acid, the chances the
Research Question
Effect of increasing the temperature of 1 moldm−3 of methanoic acid (10.00, 15.00, 20.00, 30.00, 40.00, and 50.00
±0.01 ) in the reaction of methanoic acid(aq) and sodium hydrogen carbonate solution(aq), on the rate constant of the
change in mass measured indirectly every 15 seconds for 180 seconds (±0.01s) and then using calculations and graphing
to determine the activation energy, rate of reaction, rate expression, rate constant, and the order of reaction?
Bees contain methanoic acid in their stingers (“Bee Stings - Acid or Alkali?”, n.d.). Since methanoic acid was an
acid, the best way to neutralize it was through a bicarbonate, which in my experiment was sodium hydrogen carbonate,
commonly known as baking soda. A neutralization reaction is when an acid and base react, forming water and salt, in my
case was sodium formate. This is due to H +¿¿ andOH −¿¿ions reacting together, which produces water. An equation that
was used to demonstrate this reaction was:
methanoic acid+ sodium hydrogen carbonate solution→sodium formate+carbon dioxide +water
The chemical formula of this reaction was:
HCOOH (aq) + ¿ NaHC O 3(aq) → HCOONa❑(aq) +CO 2(g ) + H 2 O❑(l ) ¿
In my experiment, I used sodium hydrogen carbonate solution because I wanted to investigate how increasing the
temperature of methanoic acid will affect the rate constant, k, which required two liquids. The rate constant was a
constant at a certain temperature. This was because as I increased the temperature of methanoic acid, the particles in the
acid gained more kinetic energy, which allowed more successful collisions with the carbonate solution. A successful
collision was when the particles collided at the correct orientation and had energy equal to or greater than the activation
energy. In order to find a numerical value of the rate, I found the percentage decrease of the original mass at 180 seconds
of the reaction and divided it by the time, which was 180 seconds.
Next, I found the rate expression, which told me whether the methanoic acid or sodium hydrogen carbonate
solution had an impact on the rate of reaction. In the equation below, reactant A was the concentration of methanoic acid
and reactant B was the concentration of sodium hydrogen carbonate solution. The concentrations were calculated and
then substituted into the equation later when I tried to find the rate constant for each temperature:
rate=k ¿
Figure 1 General rate expression equation for my investigation
The order of reaction, represented by a (methanoic acid) and b (sodium hydrogen carbonate solution), were the
powers to which the concentration of a reactant was raised in the rate expression. I found the orders of reaction by first
halving the concentrations of each reactant. Then, I conducted the same experiment with the new concentrations in
finding the changes in mass of the solution at 30 . After that, I calculated the rate of reaction of these experiments, and
used the same method as before of dividing the percentage decrease by the time. I then compared theses rates to the
reaction rate of the experiment with the original concentrations at 30 . If the rate didn’t change when I halved the
concentrations, this was zero order. First order, was when the concentrations and rates were directly proportional to each
other, so when I halved the concentration the rate of reaction halved from the original as well. Finally, when I halved the
concentration and the rate decreased by a quarter, this was second order. Once I found the orders of reaction, I found the
rate constant for each temperature. I substituted in the values of the concentrations, orders of reaction, and rate, to solve
for k for that temperature. With the rate constant I graphed a lnk vs. 1/T graph to determine the activation energy through
−E a
the Arrhenius equation: RT , as this provided me with a straight line graph. This equation also displayed the
k=Ae
−Ea
relationship between temperature (in Kelvin) and the rate constant. The slope of this equation being helped me
R
calculate the activation energy for this specific reaction.
The experimental data demonstrated the relationship between reaction rate and concentrations. The method I
chose was to investigate the change in mass of the reaction. This was because carbon dioxide was produced, which
caused the mass to decrease. This method was chosen because of a lab experiment done in class. I investigated a similar
reaction, just with different reactants, and I wanted to investigate further this in relation to the HL Kinetics topic.
I was attacked by bees as a child, where I got stung at least 30 times, which caused me to be rushed to the
hospital. Remembering this tragic experience, I wanted to investigate how I could deal with this from a chemistry view.
Hypothesis
If I increased the temperature of one moldm−3of methanoic acid, the rate constant should increase, as rate
, 2
constant is temperature-dependent. The formula to find the rate constant relied on the orders and reaction rate. Since the
orders and concentration stayed constant in my experiment, the rate constant value relied on the reaction rate.
Next, the rate of reaction should increase as well. This meant that the percentage difference from the initial mass
of the reaction was greater at higher temperatures. This was because temperature is a factor which affected the rate of
reaction. At higher temperatures, there was fast movement of methanoic acid particles which meant that the frequency of
collisions with the sodium hydrogen carbonate solution particles increased. This was due to the particles having higher
kinetic energy and entropy (measure of disorder) which allowed them to collide more often and with more speed. This
meant the particles collided with correct orientation and energy equal to or greater than the activation energy. In addition,
every 10 the rate of reaction should double, due to the general rule.
Finally, the activation energy will stay constant, showing temperature had no effect on this variable. At higher
temperatures, the activation energy was the same, there were more particles that had energy greater than or equal to that
activation energy for successful collisions. This was represented through this Maxwell-Boltzmann Distribution Curve:
Figure 2 Maxwell Boltzmann Distribution Curve for
temperature change (Sheetal, 1970)
Variables
Independent Variable:
I changed the temperature of the methanoic acid using a Heat Plate (±0.01 ) and Ice bath
Increments:
10.00, 15.00, 20.00, 30.00, 40.00, and 50.00 measured using a data harvest temperature probe with data logger (±0.01 )
Dependent Variable:
Rate constant k, (mol−3 dm 9 s−1)
How it will be measured:
I found the rate of reaction through the average percentage difference from the original mass at 180 seconds. Then, I
found the orders of reaction for each reactant. I substituted in these values to calculate for the rate constant.
Controlled Variables
Variables I How I controlled it? Why was it important?
Controlled
Concentration I used 1 moldm−3 Concentration was a factor that affected reaction rate. As the concentration of
of methanoic methanoic acid for each methanoic acid increased, the rate of reaction increased as well. This was also the
acid experiment. case if the concentration decreased. This was due to frequencies in collisions as at
higher concentrations, there were a greater proportion of particles, which increased
the chances of successful collisions. In specific, if I increased the concentration of
methanoic acid, there was a greater proportion of these particles compared to the
carbonate solution. When reacted with the solution, the chance in these particles
successfully colliding increased with greater numbers of methanoic acid particles.
Concentration I made 0.952 moldm−3 Concentration was a factor that affected the reaction rate. As the concentration of the
of sodium of sodium hydrogen sodium hydrogen carbonate solution increased, the reaction rate increased. This was
hydrogen carbonate for the whole the case if the concentration decreased as well. In specific, if I increased the
carbonate experiment. concentration of carbonate solution, there were a greater proportion of these particles
solution perdm 3compared to methanoic acid. When reacted with the acid, the chances the
