Given the thermochemical equations x2+3y2⟶2xy3δh1=−370 kj x2+2z2⟶2xz2δh2=−120 kj 2y2+z2⟶2y2zδh3=−270 kj calculate the change in enthalpy for the reaction. 4xy3+7z2⟶6y2z+4xz2

Answer : The change in enthalpy of the reaction is, -310 kJ

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. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

(3)    

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1)    

(2)    

(3)    

The expression for enthalpy of formation of will be,

Therefore, the change in enthalpy of the reaction is, -310 kJ

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j