Question

we - Google Given the following data, determine the rate law for the following reaction 2A+B+2C - products Experiment 2 [A] 0.273 0.819 0.273 0.273 [B] [C] Rate (M/s) 0.763 0.400 3.0 0.763 0.400 9.0 1.526 0.400 12.0 0.763 0.800 4 6.0 Rate = k[A][B][C] Rate = k[A][B][C] Rate = K[A] [B]*C]2 Rate = K[A]2[B] [CH Rate = k[A] [B][C] Question 16 5 pts SOCl2 gas dissociates into SO2 and Cly via the following reaction: SOCI2(e) + SO2(g) + Cl2(8) K. - 1.23 x 106 at 500K Calculate K for this reaction at 500 K. 3.00 x 108 5.05 x 105 3.34 x 10 1.23 x 106 2.46 x 106

Answer 1

Rate law
Rate = K[A]^x[B]^y[C]^z
3     = K(0.273)^x(0.763)^y(0.4)^z --------> 1
9     = K(0.819)^x(0.763)^y(0.4)^z ----------> 2
12     = K(0.273)^x(1.526)^y(0.4)^z ----------> 3
6     = K(0.273)^x(0.763)^y(0.8)^z ----------> 4
equation 1 divide by equation 2
3/9     = K(0.273)^x(0.763)^y(0.4)^z/K(0.819)^x(0.763)^y(0.4)^z
0.33   = (0.273/0.819)^x
(0.33)^1 = (0.33)^x
x   = 1
equation 1 divide by equation 3
3/12     = K(0.273)^x(0.763)^y(0.4)^z/K(0.273)^x(1.526)^y(0.4)^z
0.25     = (0.763/1.526)^y
(0.5)^2    = (0.5)^y
y = 2
equation 1 divide by equation 4
3/6     = K(0.273)^x(0.763)^y(0.4)^z/K(0.273)^x(0.763)^y(0.8)^z
0.5     = (0.4/0.8)^z
(0.5)^1   = (0.5)^z
z   = 1
Rate law
Rate = K[A]^1[B]^2[C]^1
Rate = K[A][B]^2[C] >>>>>answer
Q16.
SOCl2(g) ---------> SO2(g) + Cl2(g)
Dn = (1+1) -1 = 1
Kp = Kc(RT)^Dn
1.23*10^-6   = Kc(0.0821*500)^1
Kc   = 1.23*10^-6/(0.0821*500) = 3*10^-8
3.00*10^-8 >>>>answer
Q17.
K = 0.693/t1/2
   = 0.693/0.2
   = 3.47s^-1 >>>>answer
Q18.
molarity of CO = no of moles/volume in L
                = 3/0.5 =6M
molarity of H2O = no of moles/volume in L
                = 3/0.5 =6M
------- CO(g) + H2O(g) ------> H2(g) + CO2(g)
I------ 6--------6 -----------0 ---------0
C------ -x ------ -x --------- +x -------+x
E------ 6-x ---- 6-x --------- +x -------+x
    Kc = [H2][CO2]/[CO][H2O]
   16   = x*x/(6-x)(6-x)
   16   = (x/6-x)^2
   4    = x/6-x
4(6-x) = x
[CO2] =x = 4.8M
4.80M >>>>answer