A+B→2C When the reaction begins, the researcher records that the rate of reaction is such that 1 mole of A is consumed per minute. After making changes to the reaction, the researcher notes that 2 moles of A are consumed per minute. What change could the researcher have made to effect this change?
Answers
The researchers can do to increase the rate of consumption of mol A:
1. increase the concentration of substance B 2. raise the temperature 3. add catalyst 4. increase the pressureFurther explanationChemical reactions involve several factors, including factors such as reactants and products and external factors such as temperature and pressure
Factors that influence the speed of reaction in product formation.
Influencing factors include:
1. concentration of reactantsthe concentration of the reactants, the faster the reaction will be
2. surface areaThe form of reactant molecules in the form of powder or solids affects the speed of the reaction. For the same number of masses, the smaller particle size will make the reaction run faster because the reaction involves the larger surface area of the reactant
3. catalystCatalysts are added to help the reaction faster
Low activation energy is usually due to the addition of a catalyst. The catalyst makes activation energy more lace. The activation energy itself is a minimum of energy that must be possessed to a particle so that the collision produces a reaction
External factors that influence include:
1. PressurePressure will reduce the volume so that the reaction increases
High pressure also causes the reaction rate to be faster. The number of collisions between molecules is greater because of the distance between the molecules.
2. TemperatureHigh temperatures will speed up the reaction that occurs
the temperature is related to kinetic energy, heating makes kinetic energy increase and particles occur more often because particles move faster
For reaction
A + B ---> 2C
Reaction speed can be formulated:
[tex]\large{\boxed{\boxed{\bold{v~=~k.[A]^a[B]^b}}}[/tex]
where
v = reaction speed, M / s
k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹
a = reaction order to A
b = reaction order to B
[A] = [B] = concentration of substances
So the researchers can do to increase the rate of consumption of mol A to 2 mol / min :
1. increase the concentration of substance B 2. raise the temperature 3. add catalyst 4. increase the pressureLearn morethe factor can decrease the rate of a chemical reaction
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Keywords: reaction rate, products, reactants , the researcher
Final answer:
To double the rate of a reaction from consuming 1 mole of A per minute to 2 moles per minute, the researcher could have increased the concentration of reactants, raised the temperature, or added a catalyst.
Explanation:
If a researcher observes that the consumption rate of a reactant A doubled from 1 mole per minute to 2 moles per minute in the reaction A + B → 2C, several changes could account for this increased reaction rate. One possibility is that the concentration of either reactant A or B was increased. According to the rate law, which must be determined experimentally, the rate of the reaction can be influenced by changing the concentrations of the reactants. For instance, if the reaction is first order with respect to both A and B, as represented by the rate equation rate = k [A] [B], then doubling the concentration of either reactant would indeed double the rate.
Another possibility is increasing the temperature, which generally increases reaction rates because it results in a greater proportion of molecules having the necessary activation energy to react. Additionally, the researcher could have added a catalyst to the reaction. A catalyst provides an alternative pathway for the reaction with a lower activation energy, thus increasing the rate without being consumed in the process.
Related Questions
Calculate the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass. (Assume complete dissociation of the solute.) Tro, Nivaldo J.. Chemistry (p. 617). Pearson Education. Kindle Edition.
Answers
Answer:
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
Explanation:
Vapor pressure of water at 25 °C ,[tex]p^o= 23.8 mmHg[/tex]
Vapor pressure of the solution = [tex]p_s[/tex]
Number moles of water in 5.50% NaCl solution.In 100 gram of solution, 94.5 g of water is present.
[tex]n_1=\frac{94.5 g}{18 g/mol}=5.25 mol[/tex]
Number moles of NaCl in 5.50% NaCl solution.5.50 g of NaCl in 100 grams of solution.
[tex]n_2=\frac{5.50 g}{58.5 g/mol}=0.09401 mol [/tex]
Mole fraction of the solute = [tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{0.09401 mol}{0.09401 mol+5.25 mol}=0.01759[/tex]
The relative lowering in vapor pressure of the solution with non volatile solute is equal to the mole fraction of solute in the solution:
[tex]\frac{p^o-p_s}{p^o}=\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\frac{23.8 mmHg - p_s}{23.8 mmHg}=0.01759[/tex]
[tex]p_s=23.38 mmHg[/tex]
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
23.38 mmHg is the vapour pressure.
The pressure enforced by the vapours when in the thermodynamical equilibrium state on the system is called the vapour pressure.
How to calculate the vapour pressure?Vapour pressure [tex](p^{\circ})[/tex] of water at [tex]25 ^{\circ} \rm C[/tex] = 23.8 mmHgVapour pressure of aqueous solution = [tex](p_{s})[/tex]Step 1: Moles of water in 5.50% NaCl solution when 100 grams of solution = 94.5 g of water
[tex]\begin{aligned}\rm n &= \dfrac{94.5 \;\rm g}{18\;\rm g/mol}\\\\\\&= 5.25\;\rm mol\end{aligned}[/tex]
Step 2: Moles of NaCl in 5.50% solution when 5.50 g NaCl in 100 grams solution then,
[tex]\begin{aligned}\rm n &= \dfrac{5.50\;\rm g}{58.5\;\rm g/mol}\\\\\\&= 0.09401\;\rm mol\end{aligned}[/tex]
Step 3: Calculate the mole fraction of the solute:
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}}\\\\\rm X_{2} &= \dfrac{0.09401\;\rm mol}{0.09401 + 5.25\;\rm mol}\\\\&= 0.0175\end{aligned}[/tex]
Step 4: Calculate the vapour pressure
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}} = \dfrac{p^{\circ}-p_{s}}{p^{\circ}}\\\\0.0175 &= \dfrac{23.38 \;\text{mmHg} -p_{s}}{23.38 \;\rm mmHg}\\\\&= 23.38\;\rm mmHg\end{aligned}[/tex]
Therefore, 23.38 mmHg is the vapour pressure.
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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and heat during the process and the change of entropy of the gas. The isobaric heat capacity of the gas is Cp.m 28.253 J-K1-mol1.
Answers
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
We know that,
The relation between the [tex]C_p\text{ and }C_v[/tex] for an ideal gas are :
[tex]C_p-C_v=R[/tex]
As we are given :
[tex]C_p=28.253J/K.mole[/tex]
[tex]28.253J/K.mole-C_v=8.314J/K.mole[/tex]
[tex]C_v=19.939J/K.mole[/tex]
Now we have to calculate the entropy change of the gas.
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
[tex]\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J[/tex]
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. [tex](w=-pdV)[/tex]
(C) Heat during the process will be,
[tex]q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J[/tex]
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 25.0L tank with 4.5 mol of sulfur dioxide gas and 4.5 mol of oxygen gas at 30.°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.4 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answers
Answer:
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Explanation:
Initial Concentration of sulfur dioxide = [tex][SO_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
Initial Concentration of oxygen= [tex][O_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
[tex]2SO_2+O_2\rightleftharpoons 2SO_3[/tex]
Initially (0.18 M) (0.18 M) 0
Eq'm (0.18 -2x) (0.18 -x) 2x
Equilibrium concentration of sulfur trioxide =[tex][SO_3]=2x=\frac{1.4 mol}{25 L}=0.056 M[/tex]
x = 0.028 M
Equilibrium concentration of sulfur dioxide =[tex][SO_2]'=(0.18 -2x)=0.18 - 0.056 =0.124 M[/tex]
Equilibrium concentration of oxygen=[tex][O_2]'=(0.18 -x)=0.18 - 0.028 =0.152 M[/tex]
The expression for an equilibrium constant will be :
[tex]K_c=\frac{[SO_3]^2}{[SO_2]'^2[O_2]'}[/tex]
[tex]K_c=\frac{(0.056 M)^2}{(0.124 M)^2(0.152 M)}=1.3418\approx 1.3[/tex]
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Final answer:
To calculate the concentration equilibrium constant (Kc) for the reaction of SO2 and O2 to form SO3, find the equilibrium concentrations from the initial amounts and amount at equilibrium, then apply the equilibrium expression. The resulting Kc for the reaction at the final temperature is 0.020 when rounded to two significant digits.
Explanation:
The calculation of the concentration equilibrium constant for the reaction between sulfur dioxide and oxygen to form sulfur trioxide at a certain temperature involves using the equilibrium concentrations of reactants and products. The balanced chemical equation for the reaction is:
2 SO2(g) + O2(g) = 2 SO3(g)
Given 4.5 mol of SO2 and 4.5 mol of O2 initially in a 25.0 L tank, and 1.4 mol of SO3 at equilibrium, we can calculate the change in moles during the reaction (δ) and thus the equilibrium concentrations ([SO2], [O2], and [SO3]). The concentration equilibrium constant (Kc) is then found using the expression:
Kc = ([SO3]2)/([SO2]2 × [O2])
Through stoichiometry and equilibrium concentration calculations, the concentrations are:
[SO3] = 1.4 mol / 25.0 L = 0.056 M
[SO2] = (4.5 mol - 1.4 mol) / 25.0 L = 0.124 M (since 1 mol of SO2 is consumed for every mol of SO3 produced)
[O2] = (4.5 mol - 0.7 mol) / 25.0 L = 0.152 M (since 0.5 mol of O2 is consumed for every mol of SO3 produced)
Plugging these values into the Kc expression gives:
Kc = (0.0562) / (0.1242 × 0.152) = 0.0197
The calculated Kc at the final temperature is 0.020 (rounded to two significant digits).