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3 years ago

The standard enthalpy change for the following reaction is 873 kJ at 298 K.

Chemistry

1 answer:

Flura [38]3 years ago

8 0

Answer: - 436.5 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.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

$2KCl(s)\rightarrow 2K(s)+Cl_2(g)$ $\Delta H_1=873kJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s)$ $\Delta H_2=?$

According to the Hess’s law, if we divide the reaction by half then the $\Delta H$ will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value $\Delta H_2$ for the reaction will be:

$\Delta H_2=\frac{1}{2}\times (-873kJ)$

$\Delta H_2=-436.5kJ$

Hence, the value of $\Delta H_2$ for the reaction is -436.5 kJ.

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(A) $2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) $2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) $4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) $Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

Explanation :

(A) The given chemical reaction is :

$CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)$

The oxidation-reduction half reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}$