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DerKrebs [107]
3 years ago
13

In a coffee cup calorimeter, 1.60 g NH4NO3 was mixed with 75.0 grams of water at an initial temperature of 25.00* Celsius. After

the dissolution of the salt, the final temperature of the calorimeter contents was 23.34 * Celcius. Assuming the solution has a heat capacity of 4.18 J/g*C, and assuming no heat loss to the calorimeter, calculate the enthalpy of the solution for the dissolution of NH4NO3 in units of kJ/mol. Q
Chemistry
1 answer:
stiv31 [10]3 years ago
7 0

Answer:

-26.6kJ/mol

Explanation:

The dissolution of NH₄NO₃ is:

NH₄NO₃(aq) → NH₄⁺(aq) + NO₃⁻(aq)

To solve this question we need to find the heat released in the dissolution using the equation of coffee cup calorimeter:

Q = -m*S*ΔT

<em>Where Q is heat,</em>

<em>m is the mass of solution = 1.60g + 75.0g = 76.6g</em>

<em>S is specific heat of the calorimeter (4.18J/g°C)</em>

<em>And ΔT is change in temperature (25°C - 23.34°C = 1.66°C)</em>

<em />

The heat is:

Q = -76.6g*4.18J/g°C*1.66°C

Q = -531.5J = -0.5315kJ are released

The heat released per mole = Enthalpy for the dissolution is:

<em>Moles NH₄NO₃:</em>

1.60g * (1mol / 80.043g) = 0.0200mol

Enthalpy for the dissolution:

-0.5315kJ / 0.0200mol =

<h3>-26.6kJ/mol</h3>
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Explanation:

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That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>

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