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GREYUIT [131]
3 years ago
10

The enthalpy change for the reaction of titanium metal with gaseous iodine is given by the following thermochemical equation: 2

Ti(s) + 3 I2(g) → 2 TiI3(s) ΔH rxn = −839 kJ What is the enthalpy change for the reaction below? TiI3(s) → Ti(s) + 3/2 I2(g)
Chemistry
1 answer:
julia-pushkina [17]3 years ago
3 0

Answer : The enthalpy change for the reaction is, 419.5 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The given chemical reaction is,

2Ti(s)+3I_2(g)\rightarrow 2TiI_3(s)    \Delta H=-839kJ

Now we have to determine the enthalpy change for the reaction below:

TiI_3(s)\rightarrow Ti(s)+\frac{3}{2}I_2(g)    \Delta H'=?

By reversing and then dividing the reaction by 2, we get the enthalpy change for the reaction.

The expression will be:

\Delta H'=-\frac{(\Delta H)}{2}

\Delta H'=-\frac{(-839kJ)}{2}

\Delta H'=419.5kJ

Therefore, the enthalpy change for the reaction is, 419.5 kJ

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M( HCl) = ?

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M(KOH) = 0.620 M

Number of moles KOH:

n = M x V

n = 0.620 x 0.0127

n = 0.007874 moles of KOH

number of moles HCl :

<span>HCl + KOH = H2O + KCl
</span>
1 mole HCl ------ 1 mole KOH
<span>? mole HCl--------0.007874 moles KOH
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moles HCl = 0.007874 * 1 / 1

= 0.007874 moles of HCl

M = n / V

M = 0.007874 / <span>0.0164

</span>= 0.480 M

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