Answer:
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Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.
Reaction rate increases with concentration, as described by the rate law and explained by collision theory. As reactant concentration increases, the frequency of collision increases. The rate of gaseous reactions increases with pressure, which is, in fact, equivalent to an increase in concentration of the gas.
Answer: D) 140g
Explanation: no. of moles of NaCl = molarity X volume in litres = 2 X 1.2 = 2.4, and molar mass or mass of 1 mole of NaCl = 58.44 g, so 2.4 moles NaCl = 140.256 g
Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7B185%2C000%7D%7B2.303%20%5Ctimes%208.314%7D%20%5B%5Cfrac%7B1%7D%7B505%7D%20-%5Cfrac%7B1%7D%7B525%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%200.7289%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%2010%5E%7B0.7289%7D%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%205.356)
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356
