Question

Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished. The new equilibrium total pressure will be:_________

Answer 1

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

$V_2 = \dfrac{V_1}{\dfrac{3}{2}} ----- (1)$

also;

Thus ;

$P_1V_1 = P_2( \dfrac{V_1}{\frac{3}{2}})$

$P_1V_1 = P_2 * 2 \dfrac{V_1}{3}$

$3 P_1 V_1 = 2 P_2 V_1$

Dividing both sides by $V_1$

$3P_1 = 2P_2$

$P_2 =P_1 \dfrac{3}{2} ----- (2)$

From ;

$P_1V_1 = P_2V_2$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{2 }{3}}*V_1$

$P_2 V_2 = P_1 V_1$

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.