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deff fn [24]
3 years ago
10

Suppose that 4.9 mol no2 and 0.40 mol h2o combine and react completely. which reactant is in excess?

Chemistry
1 answer:
Elina [12.6K]3 years ago
4 0
Answer is: NO₂ (nitrogen (IV) oxide) <span>is in excess.

Balanced chemical reaction: </span>3NO₂(g) + H₂O(l) → 2HNO₃<span>(l) + NO(g).
n(</span>NO₂) = 4.9 mol.
n(H₂O) = 0.4 mol.
From chemical reaction: n(NO₂) :(H₂O) = 1 : 3.
For 0.4 mole of water, we need: 0.4 mol : n(NO₂) = 1 : 3; n(NO₂) = 1.2 mol.
Because we have more than 1.2 moles of nitrogen(IV) oxide, this molecule is in excess.
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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
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Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
The particles that make up matter do not change during a ?
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