Calculate δhrxn for the following reaction: ch4(g)+2o2(g)→co2(g)+2h2o(l) use the following reactions and given δh′s. ch4(g)+o2(g)→ch2o(g)+h2o(g), δh = -284 kj ch2o(g)+o2(g)→co2(g)+h2o(g), δh = -527 kj h2o(l)→h2o(g), δh = 44.0 kj

ΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l) : ΔH = -899 kJ

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔH °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

Delta H reaction (ΔH) is the amount of heat / heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆H ° (product) - ∑n ∆H° (reactants)

ΔH∘rxn = ΔH∘ the product (s) if ∆H ° (reactants) = 0

For the corresponding reaction we use stages of several reactions to get the desired reaction ΔH

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

Use the following reactions and given ΔH′s.

1. CH4 (g) + O2 (g) → CH2O (g) + H2O (g), ΔH = -284 kJ

2. CH2O (g) + O2 (g) → CO2 (g) + H2O (g), ΔH = -527 kJ

3. H2O (l) → H2O (g), ΔH = 44.0 kJ

Of the three ΔH known reactions we can combine to get ΔH unknown reactions

We add up the three reactions by first reversing and multiplying reaction 3 by the coefficient 2

1. CH4 (g) + O2 (g) → CH2O (g) + H2O (g), ΔH = -284 kJ

2. CH2O (g) + O2 (g) → CO2 (g) + H2O (g), ΔH = -527 kJ

3. 2H2O (g) → 2H2O (l), ΔH = -88.0 kJ

 We remove CH2O (g) because it is left and right in the reaction so the sum of the three reactions becomes:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH = -899 kJ

Delta H solution

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an exothermic reaction

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as endothermic or exothermic

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an exothermic dissolving process

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zfoods? ok, pressure is used to keep the heat inside and to keep the food perfect.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.