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At $25 ^\circ $ temperature equilibrium constant Kp for given reaction is $ = 1.8 10 ^{–7} $ Then what is the value of $ \triangle G ^\circ ? PCl_5 \rightleftharpoons PCl_3 + Cl_2 $
$ \triangle G ^\circ = – RT ln Kp $ = – 2.303RT log Kp $ = – 2.303 298 log 1.8 10^ {–7} $ = – 2.303 1.987 298 (– 6.7447) = 9197.5 Cal
The value of $ \triangle H_f ^\circ of U_3O_8 is –853.5 KJ mol^{–1} $ . Also $ \triangle H ^\circ $ for the reaction $3UO_2 + O_2 \rightarrow U_3O_8, $ is –76.00 KJ. The value of $ \triangle H_f ^\circ of UO_2 $ is approx –
To find the enthalpy of formation ($ riangle H_f^ ext{°}$) of $UO_2$, we can use the given data and the enthalpy change of the reaction. The given enthalpy changes are:
$ riangle H_f^ ext{°}(U_3O_8) = -853.5$ kJ/mol and $ riangle H^ ext{°}$ for the reaction $3 UO_2 + O_2 ightarrow U_3O_8$ is $-76.00$ kJ.
Using Hess's law, we can write the enthalpy change for the reaction as:
$ riangle H^ ext{°} = riangle H_f^ ext{°}(U_3O_8) - 3 riangle H_f^ ext{°}(UO_2) - riangle H_f^ ext{°}(O_2)$
Since the enthalpy of formation of $O_2$ in its standard state is zero, we have:
$-76.00 = -853.5 - 3 riangle H_f^ ext{°}(UO_2)$
Rearranging to solve for $ riangle H_f^ ext{°}(UO_2)$:
$3 riangle H_f^ ext{°}(UO_2) = -853.5 + 76.00$
$3 riangle H_f^ ext{°}(UO_2) = -777.5$
$ riangle H_f^ ext{°}(UO_2) = -777.5 / 3 = -259.17$ kJ/mol
Thus, the value of $ riangle H_f^ ext{°}$ of $UO_2$ is approximately $-259.17$ kJ/mol.
A gas is allowed to expand at constant pressure from a volume of 1.0 litre to 10.0 litre against an external pressure of 0.50 atm . If the gas absorbs 250 J of heat from the surroundings , what are the values of q , w and $ \triangle E $ ? ( Given 1 L atm = 101 J ) $
For this problem, we need to determine the values of $q$, $w$, and $ riangle E$.
Given data:
First, calculate the work done by the gas during expansion:
$w = -P riangle V = -P (V_f - V_i) = -0.50$ atm $(10.0 - 1.0)$ L
$w = -0.50 imes 9.0$ L atm = $-4.5$ L atm
Convert L atm to Joules:
$w = -4.5 imes 101$ J = $-454.5$ J
Rounding, $w ightarrow -455$ J
Next, calculate the change in internal energy, $ riangle E$:
$ riangle E = q + w = 250$ J $+ (-455)$ J = $-205$ J
Thus, the values are: $q = 250$ J $w = -455$ J $ riangle E = -205$ J
The enthalpy of the reaction $ H_2 O_2 ( l) \rightarrow H_2 O (l) + 1/2 O_2 ( g) $ is $ - 23.5 kcal mol ^ {-1} $ and the enthalpy of formation of $ H_2 O (l) $ is $ -68.3 kcal mol ^ {-1} $ . The enthalpy of formation of $ H_2 O_2 (l) $ is
To find the enthalpy of formation ($ riangle H_f^ ext{°}$) of $H_2O_2(l)$, we can use the given reaction and enthalpy values. The given data is:
$ riangle H^ ext{°}$ for the reaction $H_2O_2(l) ightarrow H_2O(l) + rac{1}{2}O_2(g) = -23.5$ kcal/mol
$ riangle H_f^ ext{°}(H_2O(l)) = -68.3$ kcal/mol
We need to find $ riangle H_f^ ext{°}(H_2O_2(l))$.
From the reaction, we can write:
$ riangle H^ ext{°} = riangle H_f^ ext{°}(H_2O(l)) + rac{1}{2} riangle H_f^ ext{°}(O_2(g)) - riangle H_f^ ext{°}(H_2O_2(l))$
Since $ riangle H_f^ ext{°}(O_2(g)) = 0$ (standard state), we have:
$-23.5 = -68.3 - riangle H_f^ ext{°}(H_2O_2(l))$
Rearranging to solve for $ riangle H_f^ ext{°}(H_2O_2(l))$:
$ riangle H_f^ ext{°}(H_2O_2(l)) = -68.3 + 23.5 = -44.8$ kcal/mol
Thus, the enthalpy of formation of $H_2O_2(l)$ is $-44.8$ kcal/mol.
The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is -50 kJ/ cycle . The heat absorbed by the system per cycle is -
In a cyclic process, the change in internal energy ( ext{ΔU}) of the system is zero because the system returns to its initial state. According to the first law of thermodynamics, ext{ΔU = Q - W}, where Q is the heat absorbed by the system and W is the work done by the system. Given that ext{ΔU} is zero for a cyclic process, the equation simplifies to ext{Q = W}. Since the work done by the system is -50 kJ, the heat absorbed by the system must also be -50 kJ. However, since the question asks for the heat absorbed, we consider the positive value of the work done by the system (since it is energy taken out of the system), which is 50 kJ.
$ 9.0 gm of H_2 O $ is vaporised at $ 100 ^ \circ C $ and at 1 atm pressure . If the latent heat of vapourisation of water is xJ / gm , then $ \triangle S $ is given by
Equal volumes of monoatomic and diatomic gases at same initial temperature and pressure are mixed . The ratio of specific heats of the mixture $ ( C_p / C_v ) $ will be
$ C_v = {3 \over 2} RT ; C_p = { 5 \over 2} RT $ for monoatomic gas $ C_v = { 5 \over 2} RT ; C_p = { 7 \over 2} RT $ for diatomic gas Thus for mixture of 1 mole each, $ C_v = { { 3\over 2 } RT + { 5 \over 2} RT \over 2 } and C_p= { { 5 \over 2 } RT + {7 \over 2} RT \over 2 } $ Therefore $ C_p /C_v = { 3RT \over 2 RT } = 1.5 $
The heat evolved in the combustion of benzene is given by $ C_6 H_6 + 7 { 1 \over 2 } O_2 \rightarrow 6CO_2 (g) + 3 H_2 O ( l ) ; \triangle H = -3264 kJ $ Which of the following quantities of heat energy will be evolved when $ 39_g C_6 H_6 $ are burnt
78g of benzene on combustion produces heat = – 3264.6 kJ $ \therefore 39g will produce = { - 3264.6 \over 2} = - 1632.3 kJ.$
A cylinder of gas is assumed to contain 11.2 kg of butane $ ( C_4 H_ {10} ) $. If a normal family needs 20000 kJ of energy per day. The cylinder will last : ( Given that for combustion of butane is - 2658 kJ )
First, calculate the total energy that can be provided by the 11.2 kg of butane. The molar mass of butane ( ext{C_4H_{10}}) is ext{58 g/mol}, so the number of moles in 11.2 kg is ext{11200 g / 58 g/mol = 193.1 mol}. The energy released per mole of butane is -2658 kJ, so the total energy available is ext{193.1 mol * 2658 kJ/mol = 513,679.8 kJ}. Given that a normal family needs 20000 kJ of energy per day, the number of days the cylinder will last is ext{513,679.8 kJ / 20000 kJ/day ≈ 25.68 days}. Since we are looking for an integer value, the closest option is 26 days.
In the reaction for the transition of carbon in the diamond form to carbon in the graphite form $ \triangle H $ is - 453.5 cal . This points out that
The reaction given is the transition of carbon from diamond to graphite with a negative enthalpy change (ΔH = -453.5 cal). A negative ΔH indicates that the process releases heat, meaning the product (graphite) is more stable than the reactant (diamond). Hence, graphite is more stable than diamond.
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