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Rama09 [41]
3 years ago
13

For the reaction c2h6 (g) → c2h4 (g) + h2 (g) δh° is +137 kj/mol and δs° is +120 j/k ∙ mol. this reaction is ________. question

10 options: spontaneous only at high temperature spontaneous at all temperatures spontaneous only at low temperature nonspontaneous at all temperatures
Chemistry
2 answers:
g100num [7]3 years ago
7 0
For the reaction c2h6 (g) → c2h4 (g) + h2 (g) δh° is +137 kj/mol and δs° is +120 j/k ∙ mol. This reaction is <span>spontaneous only at high temperature.
here the values of </span>δh° is +137 kj/mol and δs° is +120 kj/mol. so, both the value are wit the sign "+". when both values are positive than the reaction is not spontaneous at low temperature, the reaction is spontaneous only at high temperature.
tatyana61 [14]3 years ago
5 0

Answer: spontaneous only at high temperature

Explanation: C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)

The Gibbs free energy change is given by:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy

when \DeltaG= +ve, reaction is non spontaneous

\DeltaG= -ve, reaction is spontaneous

\DeltaG= 0, reaction is in equilibrium

\Delta H = enthalpy change = endothermic = +137 KJ

\Delta S = entropy change = +120 J/K

\Delta G=(+)-T(+)

\Delta G=(+)(-ve)

Now \Delta G= -ve when T\Delta S has more value than \Delta H

Thus reaction is spontaneous at high temperatures.

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Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
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Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

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