Thermodynamics Test

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Thermodynamics Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

$$H^{+}(aq) + NaOH (aq) \to Na^{+} + H_{2}O(l)\ \triangle H_{1} = -1390\ cals$$
$$HCN(aq) + NaOH(aq) \rightarrow Na^{+} + CN^- + H_{2}O(l)\ \triangle H_{2} = -2900\ cals$$.

What is $$\triangle H$$ value for $$HCN (aq) \to H^{+}(aq) + CN^{-}(aq)$$?

A
$$11400\ cals$$

B
$$10790\ cals$$

C
$$12500\ cals$$

D
$$9800\ cals$$

Solution
Question 2
1 / -0

A vessel contains $$100$$ litres of a liquid X. Heat is supplied to the liquid in such a fashion that, heat given equals change in enthalpy. The volume of the liquid increases by $$2$$ litres. If the external pressure is one atm, and $$202.6$$ Joules of heat were supplied, then $$[$$U $$-$$ total internal energy$$]$$ :

A
$$\Delta U = 0, \Delta H=0$$

B
$$\Delta U = +202.6J, \Delta H = +202.6$$J

C
$$\Delta U = -202.6J, \Delta H = -202.6$$J

D
$$\Delta U = 0, \Delta H = +202.6J$$

Solution
Question 3
1 / -0

Calculate the heat needed to raise the temperature of $$20\ g$$ iron from $$25^{\circ}C$$ to $$500^{\circ}C$$, if specific heat capacity of iron is $$0.45\ JK^{-1}g^{-1}$$.

A
$$4275\ J$$

B
$$225\ J$$

C
$$15.66\ J$$

D
$$2250\ J$$
Question 4
1 / -0

$$H_2(g) + Cl_2 (g) = 2HCl (g)$$;
$$\Delta{H}$$(298K) $$= -22.06 $$kcal. For this reaction, $$\Delta{U}$$ is equal to:

A
$$-22.06 + 2\times 10^{-3} \times 298 \times 2 kcal$$

B
$$-22.06 + 2 \times 298\ kcal$$

C
$$-22.06 - 2 \times 298 \times 4\ kcal$$

D
$$-22.06\ kcal$$

Solution

Enthalpy of reaction=
$$\Delta H = \Delta E + n_gR\ times T$$
$$\Delta E = \Delta H - n_gR\times T$$
$$\Delta E = -22.06- 0\times RT = -22.06Kcal$$
Question 5
1 / -0

$$9.2g$$ of toulene $${C}_{2}{H}_{8}(l)$$ is completely burnt in air. The difference in heat change at constant pressure and constant volume at $${27}^{o}C$$ is

A
$$-2.5kJ$$

B
$$+2.5kJ$$

C
Zero

D
$$-0.50kJ$$

Solution
Question 6
1 / -0

Ethyl chloride $$\left( { C }_{ 2 }{ H }_{ 5 }Cl \right) $$, is prepared by reaction of ethylene with hydrogen chloride :
$${ C }_{ 2 }{ H }_{ 4 }\left( g \right) +HCl\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 5 }Cl\left( g \right) ; \triangle H=-72.3kJ/mol$$
What is the value of $$\triangle U$$ (in kJ), if $$70\ g$$ of ethylene and $$73\ g$$ of $$HCL$$ are allowed to react at $$300\ K$$?

A
$$-69.8$$

B
$$-180.75$$

C
$$-174.5$$

D
$$-139.6$$

Solution

Mass of ethylene $$= 70 \; g$$

Molar mass of ethylene $$= 28 \; g$$

$$\therefore$$ No. of moles of ethylene $$= \cfrac{70}{28} = 2.5 \text{ moles}$$

Mass of $$HCl = 73 \; g$$

Molecular mass of $$HCl = 36.5 \; g$$

No. of moles of $$HCl = \cfrac{73}{36.5} = 2 \text{ moles}$$

Since no. of moles of $$HCl$$ are less than that of ethylene.

Hence $$HCl$$ will be limiting reagent.

$${{C}_{2}{H}_{4}}_{\left( g \right)} + {HCl}_{\left( g \right)} \longrightarrow {{C}_{2}{H}_{5}Cl}_{\left( g \right)}$$

From the reaction-

$$1$$ mole of $$HCl$$ reacts with $$1$$ mole of $${C}_{2}{H}_{4}$$ and gives $$1$$ moles of $${C}_{2}{H}_{5}Cl$$.

$$\therefore 2$$ moles of $$HCl$$ will react with $$2$$ moles of $${C}_{2}{H}_{4}$$ and will give $$2$$ mole of $${C}_{2}{H}_{5}Cl$$.

Given that $$\Delta{H} = -72.3 \; {KJ}/{mol} = -72300 \; {KJ}/{mol}$$

Given $$\Delta{H}$$ is for $$1$$ mole of product.

For $$2$$ moles,

$$\Delta{H} = 2 \times \left( -72300 \right) = 144600 \; {KJ}/{mol}$$

$$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = \left( 2 - \left( 2 + 2 \right) \right) = -2$$
As we know that,

$$\Delta{H} = \Delta{U} + \Delta{{n}_{g}}RT$$

$$\Rightarrow \Delta{U} = \Delta{H} - \Delta{{n}_{g}} RT$$

$$\Delta{U} = -144600 - \left( -2 \times 8.314 \times 300 \right) = -139.61 \; {KJ}/{mol}$$.
Question 7
1 / -0

For a certain reaction the change in enthalpy and change in entropy are 40.63 kJ $$mol^{-1}$$ and 100 $$JK^{-1}$$. What is the value of $$\Delta$$ at $$27^0C$$ and indicate whether the reaction is spontaneous or not

A
$$+10630 J\, mol^{-1}$$; spontaneous

B
$$+10630 J\, mol^{-1}$$; non spontaneous

C
$$-7990\, J\, mol^{-1}$$; spontaneous

D
$$+7900\, J\,mol^{-1}$$; spontaneous

Solution
Question 8
1 / -0

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $$CH_{3}OH_{(l)} + \dfrac {3}{2}O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)$$ At $$298\ K$$ standard Gibb's energies of formation for $$CH_{3}OH(l), H_{2}O(l)$$ and $$CO_{2}(g)$$ are $$-166.2, -237.2$$ and $$-394.4\ kJ\ mol^{-1}$$ respectively. If standard enthalpy of combustion of methanol is $$-726\ kJ\ mol^{-1}$$, efficiency of the fuel cell will be

A
$$80$$%

B
$$87$$%

C
$$90$$%

D
$$97$$%

Solution
Question 9
1 / -0

For Pt, $${H}_{2}$$($${p}_{1}atm$$) $$\left| { H }^{ \oplus}\left( 1M \right) \right| { H }_{ 2 }\left( { p }_{ 2 }\ atm \right)$$ , Pt (where, $${p}_{1}$$ and $${p}_{2}$$ are pressures) cell reaction will be spontaneous if?

A
$${p}_{1}={p}_{2}$$

B
$${p}_{1}> {p}_{2}$$

C
$${p}_{2}> {p}_{1}$$

D
$${p}_{1}=1atm$$

Solution
Question 10
1 / -0

What is free energy change $$\left( \Delta G \right) $$ when $$1.0$$ mole of water at $${ 100 }^{ o }C$$ and $$1$$ atm pressure is converted into steam at $${ 100 }^{ o }C$$ and $$2\ atm$$ pressure?

A
$$0\ cal$$

B
$$540\ cal$$

C
$$515.4\ cal$$

D
$$None\ of\ these$$
Solution
- Since, the reaction is being carried out at constant temperature $$(100^0C)$$, therefore the reaction is in thermal equilibrium. At equilibrium the value of $$\Delta{G}$$ is zero.
- This can also be proved mathematically
$$\Delta{G} = \Delta{G}^0 + RTlnQ$$
At equilibrium, $$\Delta{G}^0 = - RTlnK_{eq}$$ and $$K_{eq} = Q$$
(where $$Q$$ is the ratio of initial activity of products to reactants, and $$K_{eq}$$ is the ratio of concentrations of products to reactants at equilibrium)
Thus, $$\Delta{G} = -RTlnQ + RTlnQ = 0$$