When 10.00 g of phosphorus is burned in o2(g) to form p4o10(s), enough heat is generated to raise the temperature of 2990 g of water

When 10.00 g of phosphorus is burned in o2(g) to form p4o10(s), enough heat is generated to raise the temperature of 2990 g of water from 18.0 ∘c to 38.0 ∘c. part a calculate the heat of formation of p4o10(s) under these conditions?

Answers

Given:

Mass of P = 10.0 g

Mass of water = 2990 g

Initial temp T1 = 18 C

Final temp T2 = 38 C

To determine:

Heat of formation of P4O10

Explanation:

The reaction is:

4P + 5O₂ → P₄O₁₀

4 moles of P produces 1 mole of P4O10

Now,

moles of P burned = mass of P/atomic mass of P = 10 g/31 g.mol-1 = 0.3226 moles

Therefore, moles of P4O10 produced = 0.3226/4 = 0.0807 moles

Heat lost during the burning of P = heat gained by water

q  = mc(T2-T1)= 2990*4.18*(38-18) = 249.96 kJ

Heat of formation of P4O10 = 249.96 kJ/0.0807 moles = 3097 kJ/mol

Ans: Heat of formation = 3097 kj/mol

-3.10 × 10⁶ J/mol

Explanation:

The heat (Qw) required to raise the temperature of the water can be calculated using the following expression.

Qw = c . m . ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in the temperature

Qw = (4.186 J/g.°C) . 2990 g . (38.0°C - 18.0°C) = 2.50 × 10⁵ J

According to the law of conservation of energy, the sum of the heat released from the formation of P₄O₁₀ and the heat absorbed by the water is zero.

Qw + Qf = 0

Qf = -Qw = -2.50 × 10⁵ J

Let's consider the formation of P₄O₁₀.

4 P(s) + 5 O₂(g) → P₄O₁₀(s)

The heat of formation must be expressed per mole of P₄O₁₀. -2.50 × 10⁵ J is the heat released when 10.00 g of P react. Let's consider the following conversion factors:

The molar mass of P is 30.97 g/mol.The molar ratio of P to P₄O₁₀ is 4:1.

The heat of formation of P₄O₁₀ is:

frac{-2.50 times 10^{5}J}{10.00gP} .frac{30.97gP}{1molP} .frac{4molP}{1molP_{4}O_{10}} =-3.10 times 10^{6}J/mol

The answer is a i think

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form(n_2) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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