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SIZIF [17.4K]
3 years ago
6

Consider the reaction. c2h5oh(l)⟶ c2h6(g)+12o2(g). state what further data would be needed to obtain δrh∘ at 500∘c and 1 bar.

Chemistry
1 answer:
UkoKoshka [18]3 years ago
6 0
For this type of problem, it is essential for you to have a data on the standard heats of formation of the substances given. For elemental substances or diatomic gases, the standard heat of formation is 0. Standard means the temperature is at 0°C and pressure at 1 atm. Calculate the standard heat of reaction using:

ΔH°rxn = ∑(Stoichiometric coefficient×ΔHf of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)
Then, use this equation to find the reaction at T = 500°C and P = 1 bar:

ΔHrxn = ΔH°rxn + [∑(Stoichiometric coefficient×Cp of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)]ΔT
So, you also need the Cp or specific heat capacities of the substances.
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Read 2 more answers
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0
3 years ago
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