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nordsb [41]
3 years ago
5

2Cl(g) ⇌ Cl2(g)

Chemistry
1 answer:
Nataly [62]3 years ago
6 0

Answer:

-121.301 kJ·mol⁻¹;  57.8 J·K⁻¹mol⁻¹; 464.9 J·K⁻¹mol⁻¹; spontaneous

Explanation:

                         2Cl(g) ⇌ Cl₂(g)

ΔHf°/kJ·mol⁻¹:  121.301       0

S°/J·K⁻¹mol⁻¹:  165.190   223.0

1. ΔᵣH

ΔᵣH = products -reactants = 0 - 121.301 = -121.301 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 223.0 - 165.190 = 57.8 J·K⁻¹mol⁻¹

3. ΔS(univ)

\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}}  +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}}  -\dfrac{\Delta H_{\text{sys}}}{T}\\\\& = & 57.8 - \dfrac{-121301}{298}\\\\& = & 57.8 + 407.1\\& = & \mathbf{464.9 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}

4. Spontaneity

\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -121.301 - 298 \times 0.0578\\& = & -121.301 - 17.23\\& = &  \textbf{-138.5 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}

ΔG is negative, so the reaction is spontaneous.

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The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

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Explanation:

Sodium formate is the conjugate base of formic acid.

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K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

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K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

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HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

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