Question

Consider the reaction: X2(g)→2X(g). When a vessel initially containing 705 torr of X2 comes to equilibrium at 298 K, the equilibrium partial pressure of X is 90.0 torr . The same reaction is repeated with an initial partial pressure of 778 torr of X2 at 729 K ; the equilibrium partial pressure of X is 513 torr

Answer 1

The enthalpy on the reaction is 14.71kJ/mol

Data;

• Initial pressure = 705 torr
• Equilibrium temperature = 298K
• Equilibrium partial pressure = 90 torr
• Initial partial pressure = 778 torr
• Temperature = 729K
• equilibrium partial pressure = 513 torr

Enthalpy Change

The enthalpy change of the reaction can solved using series of step.

From the equation of the reaction;

                  $X_2 \leftrightarrow 2x$

Initial           706    0

change         -x      +2x

equilibrium   706-x   +2x

Equilibrium partial pressure of x = 90 torr

$90torr = 2x\\x = 45$

$x_2 = 705 - x = 705 - 45 = 660 torr$

Still at equilibrium;

$KP_1 = \frac{P_x^2}{P_x} = \frac{90^2}{660} = 12.27$

For the second reaction;

                    $X_2 \leftrightarrow 2x$

Initial             778    0

Change           -x      +2x

Equilibrium       733    2x

At equilibrium pressure of x

$\frac{513^2}{733} = 359.03$

To solve for the change in enthalpy

$\ln \frac{Kp_2}{Kp_1} = \frac{\delta H}{R} [\frac{T_2 - T_1}{T_2T_1} ]\\\ln \frac{359.03}{12.27} = \frac{\delta H}{8.314} [\frac{729 - 298}{758 * 298} ]\\3.376 = \frac{\delta H}{8.314} [0.01908]\\\delta H = 1471.07 J/mol$

The enthalpy on the reaction is 14.71kJ/mol

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