for the reaction 4a(g) + 3b(g) → 2c(g)
the followi
for the reaction 4a(g) + 3b(g) → 2c(g)
the following data were obtained at constant temperature:
determine the reaction order with respect to each reactant.
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.
solution: `ch\_(4)(g) + 2o\_(2)(g) to co\_(2)(g) + 2h\_(2)o(g) + "heat. (evolved)"` a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 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 the formulas for converting between degree celsius and degree fahrenheit are: °f = (°c * 9/5) + 32 °c = (°f - 32) * 5/9 to find the temperature when both are equal, we use an old algebra trick and just set of = oc and solve one of the equations. °c = (°c * 9/5) + 32 °c - (°c * 9/5) = 32 - 4/5 * °c = 32 °c = - 32 * 5/4 °c = - 40 °f = (°f * 9/5) + 32 °f - (°f * 9/5) = 32 - 4/5 * °f = 32 °f = - 32 * 5/4 °f = - 40 so the temperature when both the celsius and fahrenheit scales are the same is - 40 degrees. let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, i. given:valency of carbon is 4valency of chlorine is 1by applying criss-cross method:formula: ccl4thus, the formula for carbon tetrachloride is ccl4.ii. given:valency of element x is 1valency of oxygen is 2by applying criss-cross method:formula: x2o answer: oxygen gas is responsible for combustion. combustion is characterized as a rapid exothermic reaction that releases a significant amount of energy in the form of heat and flames as combustion reactions that can spread through a suitable medium. combustion is the outcome of dynamic or time-dependent activities between atoms, molecules, radicals, and solid boundaries at the molecular level. for combustion to occur, three conditions must be met, which are there should be a combustible material. there should be a supporter of combustion. the combustible material is heated to the ignition temperature. the presence of fuel has a significant effect. the existence of oxygen or air. the ignition temperature is kept constant. on sapling its

oxidation:
cu --> cu^2+ + 2e^-

reduction:
o\_2 + 4e^- -->2o^2- in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d`