Answer:
It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.
Answer:
Increase in the concentration of the reactants (vinegar and baking soda) leads to an increase in the rate of reaction (more volume of CO2 is evolved within a shorter time)
Explanation:
The chemical reaction between baking soda and vinegar in water is shown in the chemical reaction equation below;
NaHCO3(aq) + CH3COOH(aq) ----->CO2(g) + H2O(l) + CH3COONa(aq)
The chemical name of baking soda is sodium bicarbonate (NaHCO3) while vineager is a dilute acetic acid (CH3COOH) solution. This reaction provides a very easy set up in which we can study the effect of concentration on the rate of chemical reaction.
We must have it behind our minds that increase in the concentration of reactant species increases the rate of chemical reaction. Secondly, the rate of the reaction between baking soda and vinegar can be monitored by observing the volume of CO2 evolved and how quickly it evolves from the reaction mixture.
We can now postulate a hypothesis which states that; 'increase in the concentration of the reactants (vinegar and baking soda) leads to an increase in the rate of reaction (more volume of CO2 is evolved within a shorter time).'
If we go ahead to subject this hypothesis to experimental test, it will be confirmed to be true because a greater volume of CO2 will be evolved within a shorter time as the concentration of the reactants increases.
Exsperements? labs? chemesty? it could be a few things...
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
