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Molodets [167]
3 years ago
14

Is the following reaction spontaneous at 298 K? Answer by calculating ΔG. H2O(g) + C(s) → CO(g) + H2(g) ΔH = 131.3 kJ/mole ΔS =

134 J/mole˙K
Chemistry
2 answers:
SpyIntel [72]3 years ago
7 0

Is the following reaction spontaneous at 298 K? Answer by calculating ΔG. H2O(g) + C(s) → CO(g) + H2(g) ΔH = 131.3 kJ/mole ΔS = 134 J/mole˙K


No

Viktor [21]3 years ago
7 0

<u>Answer:</u> The correct answer is No.

<u>Explanation:</u>

We are given:

\Delta H^o_{rxn}=131.3kJ/mol=131300J/mol      (Conversion factor:  1 kJ = 1000)

\Delta S^o_{rxn}=134J/mol.K

Temperature of the reaction = 298 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

Putting values in above equation, we get:

\Delta G^o_{rxn}=131300J/mol-[(298K)\times (134J/mol.K)]=91368J/mol

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative. But, from the above calculation, the Gibbs free energy of the reaction is positive. Thus, the reaction is non-spontaneous.

Hence, the correct answer is No.

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What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv
statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

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Answer:

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