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,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

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) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

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

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3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$