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Review Langmuir Adsorption

abstract: from https://arxiv.org/pdf/2210.12150.pdf
reference:

step	inference rule	input	feed	output	step validity (as per SymPy)
1	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			3736177473 $r_{\rm adsorption} = k_{\rm adsorption} p_A [S]$	no validation is available for declarations
2	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			6955192897 $r_{\rm desorption} = k_{\rm desorption} [A_{\rm adsorption}]$	no validation is available for declarations
3	0000111104: declare assumption number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an assumption.			6783009163 $r_{\rm adsorption} = r_{\rm desorption}$	no validation is available for declarations
4	0000111556: substitute LHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	6783009163 $r_{\rm adsorption} = r_{\rm desorption}$ 3736177473 $r_{\rm adsorption} = k_{\rm adsorption} p_A [S]$		3507029294 $k_{\rm adsorption} p_A [S] = r_{\rm desorption}$	LHS diff is pdg0001966 - pdg0006850pdg0009046pdg0009067 RHS diff is -pdg0001966 + pdg0006850pdg0009046pdg0009067
5	0000111556: substitute LHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	3507029294 $k_{\rm adsorption} p_A [S] = r_{\rm desorption}$ 6955192897 $r_{\rm desorption} = k_{\rm desorption} [A_{\rm adsorption}]$		3488423948 $k_{\rm adsorption} p_A [S] = k_{\rm desorption} [A_{\rm adsorption}]$	LHS diff is pdg0001966 - pdg0006850pdg0009046pdg0009067 RHS diff is 0
6	0000111975: divide both sides by number of inputs: 1; feeds: 1; outputs: 1 Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.	3488423948 $k_{\rm adsorption} p_A [S] = k_{\rm desorption} [A_{\rm adsorption}]$	8162179726 $k_{\rm adsorption} p_A$	9562264720 $[S] = \frac{k_{\rm desorption} [A_{\rm adsorption}]}{k_{\rm adsorption} p_A}$	valid
Note about step 7: The concentration of all sites by summing the concentration of free sites $[S]$ and occupied sites
7	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			3599953931 $[S_0] = [S] + [A_{\rm adsorption}]$	no validation is available for declarations
8	0000111556: substitute LHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	3599953931 $[S_0] = [S] + [A_{\rm adsorption}]$ 9562264720 $[S] = \frac{k_{\rm desorption} [A_{\rm adsorption}]}{k_{\rm adsorption} p_A}$		4301729661 $[S_0] = \frac{[A_{\rm adsorption}]}{\left( \frac{k_{\rm adsorption}}{k_{\rm desorption}} \right) p_A} + [A_{\rm adsorption}]$	LHS diff is -pdg0003037 + pdg0009067 RHS diff is -pdg0004940
9	0000111432: factor out X number of inputs: 1; feeds: 1; outputs: 1 Factor $#1$ from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	4301729661 $[S_0] = \frac{[A_{\rm adsorption}]}{\left( \frac{k_{\rm adsorption}}{k_{\rm desorption}} \right) p_A} + [A_{\rm adsorption}]$	1268845856 $[A_{\rm adsorption}]$	2168306601 $[S_0] = \left(\frac{k_{\rm desorption}}{k_{\rm adsorption}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]$	valid
10	0000111975: divide both sides by number of inputs: 1; feeds: 1; outputs: 1 Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.	3488423948 $k_{\rm adsorption} p_A [S] = k_{\rm desorption} [A_{\rm adsorption}]$	1945487024 $p_A [S]$	5085809757 $\frac{k_{\rm adsorption}}{k_{\rm desorption}} = \frac{[A_{\rm adsorption}]}{p_A [S]}$	Algebraic error: LHS diff is pdg0006850 - pdg0006850/pdg0008379, RHS diff is pdg0004940(pdg0008379 - 1)/(pdg0009046pdg0009067)
Note about step 11: definition of equilibrium
11	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			7924063906 $K_{equilibrium} = \frac{k_{\rm adsorption}}{k_{\rm desorption}}$	no validation is available for declarations
12	0000111483: raise both sides to power number of inputs: 1; feeds: 1; outputs: 1 Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.	7924063906 $K_{equilibrium} = \frac{k_{\rm adsorption}}{k_{\rm desorption}}$	5516739892 $-1$	6240546932 $\frac{1}{K_{equilibrium}} = \frac{k_{\rm desorption}}{k_{\rm adsorption}}$	valid
13	0000111634: substitute RHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	2168306601 $[S_0] = \left(\frac{k_{\rm desorption}}{k_{\rm adsorption}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]$ 6240546932 $\frac{1}{K_{equilibrium}} = \frac{k_{\rm desorption}}{k_{\rm adsorption}}$		7517073655 $[S_0] = \left(\frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]$	LHS diff is -pdg0003037 + 1/pdg0004933 RHS diff is -pdg0004940 + pdg0008379/pdg0006850 - pdg0004940/(pdg0004933*pdg0009046)
14	0000111975: divide both sides by number of inputs: 1; feeds: 1; outputs: 1 Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.	7517073655 $[S_0] = \left(\frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]$	5426418187 $[A_{\rm adsorption}]$	6457999644 $\frac{[S_0]}{[A_{\rm adsorption}]} = \frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1$	valid
15	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			2114909846 $\theta_A = \frac{[A_{\rm adsorption}]}{[S_0]}$	no validation is available for declarations
16	0000111483: raise both sides to power number of inputs: 1; feeds: 1; outputs: 1 Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.	2114909846 $\theta_A = \frac{[A_{\rm adsorption}]}{[S_0]}$	7564010952 $-1$	8131665171 $\frac{1}{\theta_A} = \frac{[S_0]}{[A_{\rm adsorption}]}$	valid
17	0000111634: substitute RHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	6457999644 $\frac{[S_0]}{[A_{\rm adsorption}]} = \frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1$ 8131665171 $\frac{1}{\theta_A} = \frac{[S_0]}{[A_{\rm adsorption}]}$		7928111771 $\frac{1}{\theta_A} = \frac{1}{K_{\rm equilibrium} p_A} + 1$	LHS diff is 0 RHS diff is pdg0003037/pdg0004940 - 1 - 1/(pdg0004933*pdg0009046)
18	0000111886: change variable X to Y number of inputs: 1; feeds: 2; outputs: 1 Change variable $#1$ to $#2$ in Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.	7928111771 $\frac{1}{\theta_A} = \frac{1}{K_{\rm equilibrium} p_A} + 1$	7630953440 $\frac{K_{\rm equilibrium} p_A}{K_{\rm equilibrium} p_A}$ 6346902704 $1$	7267424860 $\frac{1}{\theta_A} = \frac{1+(K_{\rm equilibrium}\ p_A)}{K_{\rm equilibrium}\ p_A}$	valid
19	0000111483: raise both sides to power number of inputs: 1; feeds: 1; outputs: 1 Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.	7267424860 $\frac{1}{\theta_A} = \frac{1+(K_{\rm equilibrium}\ p_A)}{K_{\rm equilibrium}\ p_A}$	3911081515 $-1$	4689334676 $\theta_A = \frac{K_{\rm equilibrium}\ p_A}{1+K_{\rm equilibrium}\ p_A}$	valid
20	0000111341: declare final expression number of inputs: 1; feeds: 0; outputs: 0 Eq.~\ref{eq:#1} is one of the final equations.	4689334676 $\theta_A = \frac{K_{\rm equilibrium}\ p_A}{1+K_{\rm equilibrium}\ p_A}$			no validation is available for declarations

Symbols used in Langmuir Adsorption

0000004940: $ [A_{\rm adsorption}] $
0000009067: $ [S] $
0000003037: $ [S_0] $
0000001791: $ \theta_A $
0000006850: $ k_{\rm adsorption} $
0000008379: $ k_{\rm desorption} $
0000004933: $ K_{\rm equilibrium} $
0000009046: $ p_A $
0000006687: $ r_{\rm adsorption} $
0000001966: $ r_{\rm desorption} $

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0.002 seconds for compute/input_feed_output_infrule_for_step: get_expressions_from_step_id_and_expr_type HAS_INPUT9fe9b3c8-2dd0-490b-be09-1d9f1eb0b2e5
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step	inference rule	input	feed	output	step validity (as per SymPy)
1	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			3736177473 \(r_{\rm adsorption} = k_{\rm adsorption} p_A [S]\)	no validation is available for declarations
2	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			6955192897 \(r_{\rm desorption} = k_{\rm desorption} [A_{\rm adsorption}]\)	no validation is available for declarations
3	0000111104: declare assumption number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an assumption.			6783009163 \(r_{\rm adsorption} = r_{\rm desorption}\)	no validation is available for declarations
4	0000111556: substitute LHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	6783009163 \(r_{\rm adsorption} = r_{\rm desorption}\) 3736177473 \(r_{\rm adsorption} = k_{\rm adsorption} p_A [S]\)		3507029294 \(k_{\rm adsorption} p_A [S] = r_{\rm desorption}\)	LHS diff is pdg0001966 - pdg0006850pdg0009046pdg0009067 RHS diff is -pdg0001966 + pdg0006850pdg0009046pdg0009067
5	0000111556: substitute LHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	3507029294 \(k_{\rm adsorption} p_A [S] = r_{\rm desorption}\) 6955192897 \(r_{\rm desorption} = k_{\rm desorption} [A_{\rm adsorption}]\)		3488423948 \(k_{\rm adsorption} p_A [S] = k_{\rm desorption} [A_{\rm adsorption}]\)	LHS diff is pdg0001966 - pdg0006850pdg0009046pdg0009067 RHS diff is 0
6	0000111975: divide both sides by number of inputs: 1; feeds: 1; outputs: 1 Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.	3488423948 \(k_{\rm adsorption} p_A [S] = k_{\rm desorption} [A_{\rm adsorption}]\)	8162179726 \(k_{\rm adsorption} p_A\)	9562264720 \([S] = \frac{k_{\rm desorption} [A_{\rm adsorption}]}{k_{\rm adsorption} p_A}\)	valid
Note about step 7: The concentration of all sites by summing the concentration of free sites $[S]$ and occupied sites
7	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			3599953931 \([S_0] = [S] + [A_{\rm adsorption}]\)	no validation is available for declarations
8	0000111556: substitute LHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	3599953931 \([S_0] = [S] + [A_{\rm adsorption}]\) 9562264720 \([S] = \frac{k_{\rm desorption} [A_{\rm adsorption}]}{k_{\rm adsorption} p_A}\)		4301729661 \([S_0] = \frac{[A_{\rm adsorption}]}{\left( \frac{k_{\rm adsorption}}{k_{\rm desorption}} \right) p_A} + [A_{\rm adsorption}]\)	LHS diff is -pdg0003037 + pdg0009067 RHS diff is -pdg0004940
9	0000111432: factor out X number of inputs: 1; feeds: 1; outputs: 1 Factor $#1$ from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	4301729661 \([S_0] = \frac{[A_{\rm adsorption}]}{\left( \frac{k_{\rm adsorption}}{k_{\rm desorption}} \right) p_A} + [A_{\rm adsorption}]\)	1268845856 \([A_{\rm adsorption}]\)	2168306601 \([S_0] = \left(\frac{k_{\rm desorption}}{k_{\rm adsorption}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]\)	valid
10	0000111975: divide both sides by number of inputs: 1; feeds: 1; outputs: 1 Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.	3488423948 \(k_{\rm adsorption} p_A [S] = k_{\rm desorption} [A_{\rm adsorption}]\)	1945487024 \(p_A [S]\)	5085809757 \(\frac{k_{\rm adsorption}}{k_{\rm desorption}} = \frac{[A_{\rm adsorption}]}{p_A [S]}\)	Algebraic error: LHS diff is pdg0006850 - pdg0006850/pdg0008379, RHS diff is pdg0004940(pdg0008379 - 1)/(pdg0009046pdg0009067)
Note about step 11: definition of equilibrium
11	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			7924063906 \(K_{equilibrium} = \frac{k_{\rm adsorption}}{k_{\rm desorption}}\)	no validation is available for declarations
12	0000111483: raise both sides to power number of inputs: 1; feeds: 1; outputs: 1 Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.	7924063906 \(K_{equilibrium} = \frac{k_{\rm adsorption}}{k_{\rm desorption}}\)	5516739892 \(-1\)	6240546932 \(\frac{1}{K_{equilibrium}} = \frac{k_{\rm desorption}}{k_{\rm adsorption}}\)	valid
13	0000111634: substitute RHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	2168306601 \([S_0] = \left(\frac{k_{\rm desorption}}{k_{\rm adsorption}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]\) 6240546932 \(\frac{1}{K_{equilibrium}} = \frac{k_{\rm desorption}}{k_{\rm adsorption}}\)		7517073655 \([S_0] = \left(\frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]\)	LHS diff is -pdg0003037 + 1/pdg0004933 RHS diff is -pdg0004940 + pdg0008379/pdg0006850 - pdg0004940/(pdg0004933*pdg0009046)
14	0000111975: divide both sides by number of inputs: 1; feeds: 1; outputs: 1 Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.	7517073655 \([S_0] = \left(\frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1\right)[A_{\rm adsorption}]\)	5426418187 \([A_{\rm adsorption}]\)	6457999644 \(\frac{[S_0]}{[A_{\rm adsorption}]} = \frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1\)	valid
15	0000111981: declare initial expression number of inputs: 0; feeds: 0; outputs: 1 Eq.~\ref{eq:#1} is an initial equation.			2114909846 \(\theta_A = \frac{[A_{\rm adsorption}]}{[S_0]}\)	no validation is available for declarations
16	0000111483: raise both sides to power number of inputs: 1; feeds: 1; outputs: 1 Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.	2114909846 \(\theta_A = \frac{[A_{\rm adsorption}]}{[S_0]}\)	7564010952 \(-1\)	8131665171 \(\frac{1}{\theta_A} = \frac{[S_0]}{[A_{\rm adsorption}]}\)	valid
17	0000111634: substitute RHS of expr 1 into expr 2 number of inputs: 2; feeds: 0; outputs: 1 Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.	6457999644 \(\frac{[S_0]}{[A_{\rm adsorption}]} = \frac{1}{K_{\rm equilibrium}} \frac{1}{p_A} + 1\) 8131665171 \(\frac{1}{\theta_A} = \frac{[S_0]}{[A_{\rm adsorption}]}\)		7928111771 \(\frac{1}{\theta_A} = \frac{1}{K_{\rm equilibrium} p_A} + 1\)	LHS diff is 0 RHS diff is pdg0003037/pdg0004940 - 1 - 1/(pdg0004933*pdg0009046)
18	0000111886: change variable X to Y number of inputs: 1; feeds: 2; outputs: 1 Change variable $#1$ to $#2$ in Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.	7928111771 \(\frac{1}{\theta_A} = \frac{1}{K_{\rm equilibrium} p_A} + 1\)	7630953440 \(\frac{K_{\rm equilibrium} p_A}{K_{\rm equilibrium} p_A}\) 6346902704 \(1\)	7267424860 \(\frac{1}{\theta_A} = \frac{1+(K_{\rm equilibrium}\ p_A)}{K_{\rm equilibrium}\ p_A}\)	valid
19	0000111483: raise both sides to power number of inputs: 1; feeds: 1; outputs: 1 Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.	7267424860 \(\frac{1}{\theta_A} = \frac{1+(K_{\rm equilibrium}\ p_A)}{K_{\rm equilibrium}\ p_A}\)	3911081515 \(-1\)	4689334676 \(\theta_A = \frac{K_{\rm equilibrium}\ p_A}{1+K_{\rm equilibrium}\ p_A}\)	valid
20	0000111341: declare final expression number of inputs: 1; feeds: 0; outputs: 0 Eq.~\ref{eq:#1} is one of the final equations.	4689334676 \(\theta_A = \frac{K_{\rm equilibrium}\ p_A}{1+K_{\rm equilibrium}\ p_A}\)			no validation is available for declarations