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Aleksandr [31]
3 years ago
14

For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are

Chemistry
1 answer:
34kurt3 years ago
8 0

Answer:

ΔG = -6.5kJ/mol at 500K

Explanation:

We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:

ΔG = ΔH - TΔS

Computing the values in the problem:

ΔG = ?

ΔH = 2kJ/mol

T = 500K

And ΔS = 0.017kJ/(K•mol)

Replacing:

ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)

ΔG = 2kJ/mol - 8.5kJ/mol

<h3>ΔG = -6.5kJ/mol at 500K</h3>

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Answer:

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7 0
3 years ago
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Sedimentary rocks are a formed below earths surface as magma b a type of foliated igneous rock c formed by great heat d formed f
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4 years ago
When a lab report is cumulative it means that it
Kruka [31]
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3 years ago
This nonmetal is in the same group as lead.<br><br> What is the element
Andreyy89
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5 0
4 years ago
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
3 years ago
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