Δ H reaction = q / n where q: amount of heat released and n is number of moles of substance.
q = m . C . ΔT where:
m = mass of substance (g)
C = Specific heat capacity (4.18)
ΔT = change in temperature = 24.25 - 23.16 = 1.09
q = 1000 x 4.18 x 1.09 = 4556 J = 4.556 kJ
number of moles (n) = Molarity (M) x Volume (L)
= 0.185 M x 0.07 L = 0.01295 mole
Δ H = q / n = - (4.556 kJ / 0.01295 mole) = -351.8 kJ / mol
Note: it is exothermic reaction (-ve sign) i.e. temperature is raised
According to Bohr's model of the atom, the higher the orbital in which the electrons are found, the higher their energy or excitation state. Therefore, the electrons with the least amount of energy are those at the lowest orbitals, which are closer to the nucleus.
These orbitals are characterized by 4 quantum numbers, namely the principal quantum number (n), orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms). The principal quantum number reflects the distance of the electrons from the nucleus with n=1 as the orbital closest to the nucleus. Thus, according to Bohr's model, electrons in the orbital with n=1 have the lowest energy.
Answer:
Keq = [CO₂]/[O₂]
Explanation:
Step 1: Write the balanced equation for the reaction at equilibrium
C(s) + O₂(g) ⇄ CO₂(g)
Step 2: Write the expression for the equilibrium constant (Keq)
The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:
Keq = [CO₂]/[O₂]
