Which metals are called noble metals
1 answer:
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Answer:
The total change in enthalpy for the reaction is - 81533.6 J/mol
Explanation:
Given the data in the question;
Reaction;
HCl + NaOH → NaCl + H₂O
Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J
Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
so, 0.0500 moles of H₂O produced
Volume of solution = 50.mL + 50.mL = 100.0 mL
Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g
now ,
Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal
we know that; The specific heat of water(H₂O) is 4.18 J/g°C.
so we substitute
q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28
q_soln = 2842.4 + 1234.28
q_soln = 4076.68 J
Enthalpy change for the neutralization is ΔH
ΔH = -q_soln / mole of water produced
so we substitute
ΔH = -( 4076.68 J ) / 0.0500 mol
ΔH = - 81533.6 J/mol
Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol
Answer : The value of and K is, -180 kJ/mol and
Explanation :
The balanced cell reaction will be,
The half-cell reactions are:
Oxidation reaction (anode) :
Reduction reaction (cathode) :
Relationship between standard Gibbs free energy and standard electrode potential follows:
where,
= standard Gibbs free energy
F = Faraday constant = 96500 C
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = 0.93 V
Now put all the given values in the above formula, we get:
Now we have to calculate the value of 'K'.
where,
= standard Gibbs free energy = -180 kJ/mol
R = gas constant =
T = temperature = 298 K
K = equilibrium constant = ?
Now put all the given values in the above formula 1, we get:
Therefore, the value of and K is, -180 kJ/mol and