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crimeas [40]
3 years ago
14

A reaction has a standard free-energy change of -14.50 kJ mol(-3.466 kcal mol). Calculate the equilibrium constant for the react

ion at 25 °C. 5.85 K
Chemistry
1 answer:
den301095 [7]3 years ago
5 0

Answer:  The equilibrium constant for the reaction at 25 °C is 346.7

Explanation:

Formula used :

\Delta G^o=-2.303\times RT\times \log K_c

where,

\Delta G^o = standard Gibb's free energy change = -14.50kJ/mol =14500 J/mol

R = universal gas constant = 8.314 J/K/mole

T = temperature = 25^0C= (25+273)K=298 K

K_c = equilibrium constant = ?

Putting in the values we get:

-14500=-2.303\times 8.314\times 298\times \log K_c

\log K_c=2.54

K_c=antilog(2.54)=346.7

The equilibrium constant for the reaction at 25 °C is 346.7

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A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4
natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M

8 0
3 years ago
What are the ingredients in Swedish fish
Juli2301 [7.4K]
INGREDIENTS:SUGAR, INVERT SUGAR, CORN SYRUP, MODIFIED CORN STARCH, CITRIC ACID, WHITE MINERAL OIL, NATURAL AND ARTIFICIAL FLAVOR, RED 40, AND CARNABUBA WAX.
7 0
3 years ago
ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea
kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

<u>Step 1: </u>Data given

H2(g) + I2(g) ⇌ 2HI(g)     ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

<u>Step 2</u>: Calculate ΔG

ΔG = ΔG° + RTln Q  

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q  

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)  

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

4 0
3 years ago
T
svet-max [94.6K]

Explanation:

(1) The nucleus is positive and the electron cloud is positive.

(2) The nucleus is positive and the electron cloud is negative.

(3) The nucleus is negative and the electron cloud is positive.

(4) The nucleus is negative and the electron cloud is negative

4 0
3 years ago
N2-3H2 → 2NH3
aniked [119]

Answer:

6 mols H2 are needed

Explanation:

N2 = 28.01g/mol

H2 = 2.02g/mol

\frac{2 mol N_{2} }{1} * \frac{3 mol H_{2}  }{ 1 mol N_{2} } = 6 mol H2

8 0
3 years ago
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