What is the activation energy for the isomerization of methyl isocyanide?
Rate Constant
Temperature
1
4.30 x 10-3 min-1
472 K
4
7
5.54 x 10-2 min-1
503 K
+/-

18

The answer to the question is d

Following are the  advantages of the parametric equations: 1) they can be used to graph curves that are not functions, like the unit circle.2) the parameter can be used to represent the quantity that is  useful to  provide us with better information about the graph.3)  it is useful to convert between rectangular equations and parametric equations.with specific reference to chemistry,  parametric equations are employ for solving hydrogen and hydrogen-like ions, quantum mechanically. herein, cartesian coordinates are converted in spherical polar coordinates. the solution obtained, when plotted on spherical polar graph paper, provides visualization of atomic orbital such as s,p,d and f. 

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj