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The rate of formation of a product depends on the the concentrations of the reactants.If products A and B produce product C, a general equation for the formation of C is of the kind rate = k*[A]^a * [B]^b[ ] is used for the concentration of each compound.When k, m and n are unknown, chemists run lab trials to calculate them.. You have these data from 3 trialsTrial [A] [B] Rate (M) (M) (M/s)1 0.50 0.010 3.0×10−3 2 0.50 0.020 6.0×10−3 3 1.00 0.010 1.2×10−2Trials 1 and 2 are run at constant [A] which permits to calculate the exponent b, in this wayrate 1 = 3.0 * 10^ -3 = k [A1]^a * [B1]^brate 2 = 6.0*10^-3 = k [A2]^a * [B2]^b divide rate / rate 1 => 2 = [B1]^nb/ [B2]^b[B1] = 0.010 and [B2] = 0.020 => 6.0 / 3.0 =( 0.020 / 0.010)^b => 2 = 2^b => b = 1 In the same way trials 1 and 3, which were run at constan [B], are used to calculate arate 3 / rate 1 = 12 / 3.0 = (1.0)^a / (0.5)^a => 4 = 2^a => 2^2 = 2^a => 2 = aNow use any of the data to find kWith the second trial: rate = 6*10^-3 m/s = k (0.5)^2 * (0.02) =>k = 6.0*10^-3 M/s / (0.05 M^3) = 0.12 M^-2 s^-1With the calcualted values of k, a and b you use the formula of the rate with the concentrations givenrate = k[A]^2*[B] = 0.12 M^-2 s^-1 * (0.50M)^2 * (0.075M) = 0.0045 M/s = 4.5*10^=3 M/s 4.5 * 10^-3 M/s
2.3*10^-2 M/s IS THE ANWSER
The initial rate for the formation of C : v = 2.194.10⁻² M/sFurther explanationThe reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.Can be formulated:Reaction: aA ---> bBorA = reagentB = productv = reaction ratet = reaction timeFor A + B reactions ---> C + DReaction speed can be formulated: wherev = reaction speed, M / sk = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹a = reaction order to Ab = reaction order to B[A] = [B] = concentration of substancesReactions that occur:A + 2B --> CSo the reaction speed equation:v = k. [A]ᵃ [B]ᵇ1. look for the reaction order awe look for the same data [B], namely data 1 and 3 4 = 2ᵃa = 22. look for the reaction order bwe look for the same data [A], namely data 1 and 22 = 2ᵇb = 1So the rate reactionv = k .[A]²[B]To find K , use data 11.4.10⁻³ = k [0.2]²[0.03]k = 1.17 M⁻¹S⁻¹The initial rate for the formation of c at 25 ∘c, if [A]=0.50 and [B]=0.075v = 1.17 [0.5]²[0.075]v = 2.194.10⁻² M/sLearn morethe factor can decrease the rate of a chemical reactionincrease the rate of a chemical reactionWhich of the following does not influence the effectiveness of a detergentKeywords: reaction rate, reaction order, molar concentration, products, reactants
The initial rate for the formation of C at 25°C, if [A]=0.50M and [B]=0.075M is 0.022 M/sFurther explanationCalculate the initial rate for the formation of C at 25°C, if [A]=0.50M and [B]=0.075M. Express your answer to two significant figures and include the appropriate units. Consider the reactionA+2B → Cwhose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:Trial [A] [B] Rate (M) (M) (M/s)1 0.20 0.030 1.4 x 10 - 3 2 0.20 0.060 2.9 x 10 - 33 0.40 0.030 5.8 x 10 - 3We will take a look on 1st and 3rd data.because have units of so(2 significant values)Learn moreLearn more about units Learn more about the initial rate Learn more about two significant figures Answer detailsGrade: 9Subject: chemistryChapter: chemical kineticsKeywords: the initial rate, formation, c, time, M
The question is incomplete, here is the complete question:Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reactionA + 2B ⇔ Cwhose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:The table is attached below as an image. The initial rate for the formation of C at 25°C is Explanation:Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.For the given chemical equation:Rate law expression for the reaction:where,a = order with respect to Ab = order with respect to BExpression for rate law for first trial: ....(1)Expression for rate law for second trial:....(2)Expression for rate law for third trial:....(3)Dividing 2 by 1, we get:Dividing 3 by 1, we get:Thus, the rate law becomes: ......(4)Now, calculating the value of 'k' by using any expression.Putting values in equation 1, we get:Calculating the initial rate of formation of C by using equation 4, we get:[A] = 0.50 M[B] = 0.075 MPutting values in equation 4, we get:Hence, the initial rate for the formation of C at 25°C is
Rate = k * [A]^2 * [B]^1 Use the data from any trial to calculate k. k = (rate)/([A]^2 * [B]^1) E.g., for Trial 1, we have rate = 3.0×10−3 M/s [A] = 0.50 M [B] = 0.010 M Plug those numbers in and crank out the answer. Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M rate = k * [A]^2 * [B]^1 k = calculated value [A] = 0.50 M [B] = 0.075 M
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:rate = k*[A]^m * [B]^nWhere the symbol [ ] is the concentration of each compound.Then, plus the concentrations of compounds A and B you need k, m and n.Normally you run controled trials in lab which permit to calculate k, m and n .Here the data obtained in the lab are:Trial [A] [B] Rate (M) (M) (M/s)1 0.50 0.010 3.0×10−3 2 0.50 0.020 6.0×10−3 3 1.000 .010 1.2×10−2Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this wayrate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^nrate 2 = 6.0*10^-3 = k [A2]^m * [B2]^ndivide rate / rate 1 => 2 = [B1]^n / [B2]^n[B1] = 0.010 and [B2] = 0.020 => 6.0 / 3.0 =( 0.020 / 0.010)^n =>2 = 2^n => n = 1 Given that for data 1 and 3 [B] is the same, you use those data to find mrate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>4 = 2^m => m = 2Now use any of the data to find kWith the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1Now that you have k, m and n you can use the formula of the rate with the concentrations givenrate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s 4.5 * 10^-3 m/s
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