Kp=Kc(RT)^(Δn) where R, related to concentration, is 0.08206 ~= 0.0821, Kp is pressure in atm of ?, Kc is equilibrium constant (unitless), T is temperature in Kelvin, delta n is moles of products(g) minus moles of reactants(g).

Kc or Kp=([product 1]^coefficient*[product 2]^coefficient)/([reactant 1]^coefficient*[reactant 2]^coefficient) where products and reactants are only factored in if they are gaseous I guess.

Kc_reverse=1/Kc_forward

Kc is the same, even if the concentration changes.

#2
* not an answer: 1=Kc(0.0821*273)^1; 1=Kc(22.4133); Kc=0.044616366.
* not an answer: 1=Kc(0.0821*273)^0; 1=Kc.
* no: Kc=(1^1*1^1)/(1^2)
* Kc_reverse=1/38,000=0.000 026 316

#3
* not an answer: 1=0.000 000 000 83(0.0821*298)^(-1); 1=/=<1
* 8.3e-10=[NO2]^2/([N2][O2]^2); (How to apply exponents to parantheses: (2*4)^2=(2^2*4^2)=4*16=64=(8)^2=64); 8.3e-10=((5x)^2)/([N2]*x^2); 8.3e-10=(25*x^2)/([N2]*x^2); 8.3e-10=25/[N2]; 8.3e-10[N2]=25; [N2]=3.012 048 193e10
CHECK

Kp=61(0.0821*400)^(-2); Kp=61(32.84)^(-2); Kp=0.056561841

9^2/6^2*0.45^1; 81/36*0.45; 81/16.2=5

Can equilibrium constant be found from equilibrium concentrations? 1.6^2/0.14*0.39; 2.56/0.0546=46.886446689
CHECK

10.
  butane <---> isobutane
I 5 atm        0 atm
C -x           +x
E 5-x          x
25 (which is Kp) = x/(5-x); 25(5-x)=x; 125-25x=x; 26x=125; x=4.8076923076
E 5-4.8076     4.8076
E 0.192308     4.8076

11. It is not 0.58
E is I+C. FOIL=First, Outside, Inside, Last
(a+b)(a+b)
F aa 00
O ab 01
I ba 10
L bb 11
  2H20(g) + 2SO2(g) <---> 2H2S(g) + 3O2(g)
I 2.8 M     2.6 M         0 M       0 M
C -2x       -2x           +2x       +3x
E 2.8-2x    2.6-2x        2x        3x
0.000 001 3 (which is Kc) = (2x)^2(3x)^3/(2.8-2x)^2(2.6-2x)^2;
(2.8-2x)^2 = (2.8-2x)(2.8-2x) = (4x^2-11.2x+7.84)
(2.6-2x)^2 = (2.6-2x)(2.6-2x) = 6.76-5.2x-5.2x+4x^2=
4x^2*27x^3/(4x^2-11.2x+7.84)(6.76-5.2x-5.2x+4x^2)=
4x^2-11.2x+7.84
a    b     c
4x^2-10.4x+6.76
a    d     e=
1aa 16x^4
2ad -41.6x^3
3ae 27.04x^2
4ba -44.8x^3
5bd 116.48x^2
6be -75.712x
7ca 31.36x^2
8cd -81.536x
9ce 52.9984=
4x^2*27x^3/16x^4-41.6x^3+27.04x^2-44.8x^3+116.48x^2-75.712x+31.36x^2-81.536x+52.9984=0.0000013; x=0.0557547=no.
  (1/2)H20(g) + (1/2)SO2(g) <---> (1/2)H2S(g) + (1/3)O2(g)
I 2.8 M         2.6 M             0 M           0 M       
C -(1/2)x       -(1/2)x           +(1/2)x       +(1/3)x   
E 2.8-0.5x      2.6-0.5x          0.5x          0.3333333x
(0.5x)^(1/2)(0.3333333x)^(1/3)/(2.8-0.5x)^(1/2)(2.6-0.5x)^(1/2); 
0.707106781x^(1/2)*0.693361274x^(1/3)/...=
0.707106781*sqrt(x)*0.693361274*cubert(x)/sqrt(2.8-0.5x)*sqrt(2.6-0.5x)=no.
0.000 001 3=4x^2*27x^3/(2.8-2x)^2*(2.6-2x)^2; 108x^5/(2.8-2x)^2*(2.6-2x)^2=0.000 0013. For some dumbass reason ("Kc is so small compared to the initial concentrations [blah blah blah]" --https://socratic.org/questions/a-reaction-mixture-initially-contains-2-8-m-h-2o-and-2-6-m-so-2-how-do-you-deter) you remove the -2xs. 108x^5/(2.8)^2*(2.6)^2=108x^5/7.84*6.76; 108x^5/52.9984=0.0000013; 0.000068898=108x^5; 5root(x^5)=5root(0.000 000 638); x=0.057671884, so 2x=0.115343768

12. not 3.02
Kp=x/(0.000 222)(0.000 222)? 149,000,000=x/0.000 000 049; x=7.343316
CHECK

13. to reactants.

14. Kp=([product 1]^coefficient*[product 2]^coefficient)/([reactant 1]^coefficient*[reactant 2]^coefficient). 22,500=1/[CO]*(0.52)^2; 22,500=1/0.2704x. 6084x=1; x=0.000 164 366

17. Kp=Kc(RT)^n; 41=Kc(0.0821*400)^(-2); 41=Kc(32.84)^(-2); 41=Kc*0.000 927 243; Kc=44,217.0896

18. Delta number of moles

20. Kc will not decrease.