Chemistry, 03.12.2019 00:31 droidd133

Given thath2(g) + f2(g) -> 2hf(g) => ∆h = -546.6 kj . mol-12h2(g) + o2(g) -> 2h20(l) => ∆h = -571.6 kj. mol-1

calculate the value of ∆h for: 2f2(g) + 2h20(l) -> 4hf(g) + o2(g)

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Answer from: Kurlyash

The $\Delta H^o_{rxn}$ for the reaction is -521.6 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of fluorine and water follows:

$2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $H_2(g)+F_2(g)\rightarrow 2HF(g)$ $\Delta H_1=-546.6kJ$ ( × 2)

(2) $H_2(g)+O_2(g)\rightarrow 2H_2O(g)$ $\Delta H_2=-571.6kJ$

The expression for enthalpy of reaction follows:

$\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-546.6))+(1\times (571.6))]=-521.6kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -521.6 kJ.

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

answer; i do believe the correct answer is (river); /// good luck

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Given thath2(g) + f2(g) -> 2hf(g) => ∆h = -546.6 kj . mol-12h2(g) + o2(g) -> 2h20(l) =&g...