Answer to Question #42344 in Physical Chemistry for qwerty




 Consider the following reactions at 300 K.
A → B (uncatalysed reaction)
A →B (catalyst reaction)
The activation energy is lowered by 8.314 KJ mol–1 for the catalysed reaction. How many times the rate of this catalysed reaction greater than that of uncatalysed reaction ?
(1) 15 times
(2) 18 times
(3) 22 times
(4) 28 times
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hi there!we are asked to determine the reaction order of a and b using the given data.recall that the rate law only focuses on the reactant concentrations and has a general form of:rate= kaxbyk = rate constanta & b = reactantsx & y = reactant ordersreactant ain order to calculate the order of the reaction with respect to a, we must look for the experimental data where the concentration of a changes while the concentration of b remains the same.trial[a] (mol/l)[b] (mol/l)initial rate (m/s)10.1000.100520.3000.1004530.1000.2001040.3000.20090the rate laws are:rate= kaxbytrial 1:5 m/s= k0.100x0.100ytrial 2:45 m/s= k0.300x0.100ysolve for x using the ratio of the two rates:***the larger rate should be on the numeratorrate 2rate 1=45 m/s5 m/s= k0.300x0.100yk0.100x0.100y• for reactions at a constant temperature, the rate constant remains the same. thus, it could be canceled out.• since the reaction order ‘y’ of b is also the same, it could also be canceled out.45 m/s5 m/s= k0.300x0.100yk0.100x0.100y9 = 3xlog 9=x·log3log 9log3=x·log3log3x=2the reaction order, x, with respect to a is 2. reactant bin order to calculate the order of the reaction with respect to a, we must look for the experimental data where the concentration of b changes while the concentration of a remains the same.trial[a] (mol/l)[b] (mol/l)initial rate (m/s)10.1000.100520.3000.1004530.1000.2001040.3000.20090the rate laws are:rate= kaxbytrial 1:5 m/s= k0.100x0.100ytrial 3:10 m/s= k0.100x0.200ysolve for x using the ratio of the two rates:***the larger rate should be on the numeratorrate 3rate 1=10 m/s5 m/s= k0.100x0.200yk0.100x0.100y• for reactions at a constant temperature, the rate constant remains the same. thus, it could be canceled out.• since the reaction order ‘x’ of a is also the same, it could also be canceled out.10 m/s5 m/s= k0.100x0.200yk0.100x0.100y2 = 2ylog 2=y·log2log 2log2=y·log2log2y=1the reaction order, y, with respect to b is 1. kclo3(s) — heat→ kcl (s) + o2(g)      reactant        productk        1                   1cl       1                   1o       3                    2[readmore]2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g)      reactant        productk        2                   2cl       2                   2o       6                    6the balanced equation for the reaction is 2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g) the q value for the nernst equation can be determined by  the overall reaction of the cell notation. recall that q is the reaction quotient which can be calculated as:determine the overall reaction using the mnemonic for cell notation, abc:step 1. determine the overall reaction and find the anode and cathode. the anode (one being oxidized) and cathode (one being reduced). using the mnemonic, it appears that the anode is zn(s) while the cathode is cu(s).step 2. write the equation where the anode, zn (s), is being oxidized to to zn2+ (from zn(no3)2) and the cathode, cu (s), is produced by reduction of cu2+ (from cu(no3)2zn (s) - anode; cu (s) - cathode zn (s) + cu2+ (aq) → cu (s) + zn2+ (aq)step 3. setup the q expression using the concentration of zn2+ and cu2+ provided. note that (s) and (l) are not included in q or keq expressionsstep 4. plugin q to lnq ln(q) = ln(1) = 0 answer is 0 setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m generaloraverage rate is the change in the concentration of a compound over a period of time.the average rate the reactants and products in a reaction are equal when the coefficient for each is 1.the rate for the reactants are negative. this is because the reactants are being consumed in the reaction. the rate for the product c is positive because it is being produced in the reaction.for a reaction where the coefficients of the compounds are different, the rate is multiplied by the reciprocal of the coefficient.example: 2a + b → 3cthe given reaction is:3o2(g) → 2o3(g)­given the rate of o2 consumption is -2.17x10-5 mol/l∙snote that value is negative to indicate that it is decreasing as the reaction proceeds.rearranging the rate equation to give the reaction rate is terms of [o3]: [readmore]typed answer2no + o2 → 2no2rate=knomo2n2nd order in no → m = 23rd order overall → m + n = 3so 2 + n = 3n = 1rate=k[no]2[o2] we are being asked to classify the following reactions based on the following categories:combination/synthesis → multiple reactants combine to form a single productw/o change in oxidation numbersa + b → abdecomposition →reactant splits into multiple products with no change in oxidation numbersab → a + bcombustion reaction: reaction with oxygen that produces carbon dioxide and waterhydrocarbon + o2 → co2 + h2osingle displacement →single element displaces another element in a compoundab + c → ac + b[readmore]let’s analyze the given reactionsa. fe2(co3)2 (s) → fe2o3(s) + 2co2(g) break down of fe2(co3)2(s) to its component ions → decomposition reactionanswer: this is a decomposition reaction.b. 2 c6h11oh(l) + 17 o2(g) → 12co2 (g) + 12h2o(l) reaction with oxygen that produces carbon dioxide and water → combustion reactionanswer: this is a combustion reaction.c. p4o10(s) + 6h2o(l) → 4h3po4(aq) two or more substances combine to form a single compound → combination reactionanswer: this is a combination reaction.d. 2c(s) + mno2(s) → mn(s) + 2co(g) substitute one element for another element in a compound →single-displacement reaction answer: this is a single-displacement reactione. 2naclo3 (s) → 2nacl(s) + 3o2(g)break down of compound into its simpler substances → decomposition reactionanswer: this is a decomposition reaction heat lost by the vessel and water= [150 x 4.2 x (30 - 5) j + 100 x 0.4 x (30-5)] j = (150 x 4.2 x 25 + 100 x 0.4 x 25) j = (15,750 + 1000) j= 16,750 j heat gained by ice to change into water at 0 °c = m x specific latent heat of fusion of ice = {m x 336} heat gained by water to come to final temperature (5 °c) = m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m if there is no loss of heat to the surroundings, according to the principle of calorimetry: heat gained by ice to melt and come to final temperature = heat lost by vessel and water or 336m + 21m= 16,750 or 357m = 16,750 or m = 16750/357 = 46.92 ?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v according to ncert answer would be option (1)and

but according to general chemistry answer would be (3)andoption 1)and





option 2)andoption 3)andoption 4)and cao