FINISHED_Final_Notebook_Jones

B. decreasing the concentration of C​ 6​
H​6
C. increasing the concentration of HBr
D. decreasing the temperature
Explanation: ‘A’ would be the answer because the greater the concentration of one of the reactants,
the higher the possibility of collision. The other answers are wrong either because they have the
opposite effect, or deal with the product, not the reactants. Decreasing the concentration of C​ 6​
H​6
would simply decrease the reaction rate because there is less availability of reactants to make
products. Increasing the concentration of HBr would have absolutely no effect on the reaction rate of
C​6​H​6​
+ Br​2​ because it is a product, not a reactant. Lastly, decreasing the temperature would have the
opposite of the desired effect, because the lower the kinetic energy of the reactants, the slower they
move, thus the less the reactants ability to collide and make product.
2CO + O​2​
→ 2CO​2
6. If the above reaction takes place inside a sealed reaction chamber, then which of these procedures
will cause a decrease in the rate of reaction?
A. raising the temperature of the reaction chamber
B. increasing the volume inside the reaction chamber
C. removing the CO​2​ as it is formed
D. adding more CO to the reaction chamber
Explanation: Increasing the volume would decrease the rate of reaction because the greater the
space available for particles to occupy, the less the concentration, and thus there is a less likely
chance of collision, which subsequently results in a decrease in the rate of reaction. That is why ‘B’ is
the answer. Raising the temperature of the reaction chamber would increase the rate of reaction
because if the temperature of the reactants is increased, that means there is a greater kinetic energy,
which means greater chance for collision, which means a higher reaction rate (which is the opposite
of the desired effect). Removing the CO​ 2​
as it was formed would just leave more room for the product
of the reactants, and although it means the reactants can continue to make product, it does not mean
the reaction rate will increase.
7. A catalyst can speed up the rate of a given chemical reaction by
A. increasing the equilibrium constant in favor of products.
B. lowering the activation energy required for the reaction to occur.
C. raising the temperature at which the reaction occurs.
D. increasing the pressure of reactants, thus favoring products.
Explanation: ​Catalysts​ are substances that decrease the activation energy of a reaction, and as a
result increase reaction rate. This clearly demonstrates why ‘B’ is the answer. “Increasing the
equilibrium constant in favor in of products” is not the answer because an ​equilibrium constant ​is a
quotient, or ratio, of the reaction when it hits equilibrium. It is really a measurement of the ratio of
elements involved in equilibrium, rather than anything that actually affects it. “Raising the temperature
at which the reaction occurs” implies raising the activation energy, which has the exact opposite effect
of a catalyst, and accomplishes the complete opposite task. “Increasing the pressure of reactants,
thus favoring products” is not what a catalyst does, simple as that. Catalysts merely lower the
activation energy of a reaction, they do not affect, whatsoever, the conditions under which the
reaction is subjected to.
8. Which reaction diagram shows the effect of using the appropriate catalyst in a chemical reaction?
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