Reaction involving gold have been of particular interest to a chemist- Consider the following reactions- AuOH3-4HCl - HAuCl4-3H2O- - H - -28kCal AuOH3-4HBr - HAuBr4-3H2O- - H - -36-8kCalIn an experiment there was an absorption of 0-44kCal when one mole of HAuBr4 was mixed with 4 moles of HCl- What is the percentage conversion of HAuBr4 into HAuCl4-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Heat of Reaction

Reaction invo...

Question

Reaction involving gold have been of particular interest to a chemist. Consider the following reactions,
$A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O$ , $Δ H = - 28 k C a l$
$A u (O H)_{3} + 4 H B r \to H A u B r_{4} + 3 H_{2} O$ , $Δ H = - 36.8 k C a l$
In an experiment there was an absorption of $0.44 k C a l$ when one mole of $H A u B r_{4}$ was mixed with $4 m o l e s$ of $H C l$ . What is the percentage conversion of $H A u B r_{4}$ into $H A u C l_{4}$ ?

0.5 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.6 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

50 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $5 %$
$A u (O H)_{3} + 4 H C L ⟶ H A u C L_{4} + 3 H_{2} O, △ H_{1} = - 28 k C a l$
$A u (O H)_{3} + 4 H B r ⟶ H A u B r_{4} + 3 H_{2} O, △ H = - 36.8 k C a l$

$H A u B r_{4} ⟶ 4 H C L ⟶ H A u C l_{4} + 4 H B r$
$△ H = △ H_{1} - △ H_{2}$
$= - 28 + 36.8$
$= 8.8 k C a l$

Heat $. O .44 k C a l$
% reaction $\frac{0.44}{8.8} \times \frac{100}{10} = 5$ %

Suggest Corrections

Similar questions

Q. Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,

A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O, △ H = - 28 k c a l

A u (O H)_{3} + 4 H B r \to H A u B r_{4} + 3 H_{2} O, △ H = - 36.8 k c a l

In an experiment, there was an absorption of 0.44 kcal when one mole of

H A u B r_{4}

was mixed with 4 moles of HCl. Then, the fraction of

H A u B r_{4}

converted into

H A u C l_{4}

(percentage conversion)

Q. Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,

A u (O H)_{3} + 4 H C l \to H A u C l_{4} + 3 H_{2} O, △ H = - 28 k c a l

A u (O H)_{3} + 4 H B r \to H A u B r_{4} + 3 H_{2} O, △ H = - 36.8 k c a l

In an experiment, there was an absorption of 0.44 kcal when one mole of

H A u B r_{4}

was mixed with 4 moles of HCl. Then, the fraction of

H A u B r_{4}

converted into

H A u C l_{4}

(percentage conversion)

For the conversion of 1 mole of $S O_{2} (g)$ into $S O_{3} (g)$ the enthalpy of reaction at constant volume, $Δ U$ at $298 K$ is $- 97.027 K J$ . What is the enthalpy of reaction, $Δ H$ at constant pressure?

Q. At

25^{\circ} C

1

mole of

M g S O_{4}

was dissolved in water, the heat evolved was found to be

91.2 kJ

. One mole of

M g S O_{4} .7 H_{2} O

on dissolution gives a solution of the same composition accompanied by an absorption of

13.8 kJ

. The enthalpy of hydration, i.e.,

Δ H

for the reaction

M g S O_{4} (s) + 7 H_{2} O (l) \to M g S O_{4} .7 H_{2} O (s)

is:

Q. Calculate the

Δ H

for conversion of C (diamond) to C (graphite) when the following reactions are given:

C (d i a m o n d) + O_{2} \to C O_{2} (g); Δ H = - 94.5 K c a l

C (g r a p h i t e) + O_{2} \to C O_{2} (g); Δ H = - 94 K c a l

Join BYJU'S Learning Program

The correct option is B 5%Au(OH)3+4HCL⟶HAuCL4+3H2O,△H1=−28kCalAu(OH)3+4HBr⟶HAuBr4+3H2O,△H=−36.8kCalHAuBr4⟶4HCL⟶HAuCl4+4HBr△H=△H1−△H2=−28+36.8=8.8kCalHeat .O.44kCal% reaction 0.448.8×10010=5%