Self Studies

Thermodynamics Test - 46

Result Self Studies

Thermodynamics Test - 46
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    Born-Haber cycle can be used to estimate :
    Solution

  • Question 2
    1 / -0
    Which one of the following gases possesses the largest internal energy?
    Solution

  • Question 3
    1 / -0
    Which of the following relationship is not correct for the relation between $$\Delta H$$ and $$\Delta  U$$?
    Solution
    We know,

    $$\Delta H=\Delta U+\Delta n_{g}RT$$

    When $$\Delta n_{g} = 0$$ then $$\Delta H=\Delta U$$
  • Question 4
    1 / -0
    For the electrochemical cell, $$ \mathrm{M} | \mathrm{M}^{+} \| \mathrm{X}^{-} \mid \mathrm{X}, $$
    $$ E_{M}^{o}+/ M=0 \cdot 44 V $$ and $$ E_{x / x^-}^{\circ}=0 \cdot 33 V $$From these data one can deduce that
    Solution
    $$ \mathrm{M}^{+}+e^{-} \rightarrow \mathrm{M} ; \quad \mathrm{E}^{\circ}=+0 \cdot 44 \mathrm{V} \dots(i)$$
    $$ X+e^{-} \rightarrow X^{-} ; \quad E^{\circ}=+0.33 \mathrm{V} \dots(ii)$$
    Subtracing Equ. (ii) from Eqn. (i), we have
    $$M^{+}-X \rightarrow M-X^{-} ; \quad E^{\circ}=+0 \cdot 11 V$$
    or $$ \quad M^{+}+X^{-} \rightarrow M+X ; \quad E^{\circ}=+0 \cdot 11 V $$
    i.e., $$ \quad M^{+}+X^{-} \rightarrow M+X $$ is the spontaneous reaction.
  • Question 5
    1 / -0
    Hess's law is applicable for the determination of heat of
    Solution
    According to this law, the total enthalpy change is independent of the intermediate steps involved in the change. It depends only on initial and final values of enthalpy change.
  • Question 6
    1 / -0
    The relationship between enthalpy and internal energy change is
    Solution
    • The change in the enthalpy of the system during a chemical reaction is equal to the change in its internal energy plus the change in the product of the pressure times the volume of the system. i.e

    • $$\Delta H=\Delta E+P\Delta V$$

    Hence option B is correct answer.

  • Question 7
    1 / -0
    Consider the following processes.
    $$\Delta H (kJ/mol)$$
    $$1/2A\rightarrow B$$+150
    $$3B\rightarrow 2C+D$$-125
    $$E+A\rightarrow 2D$$+350
    For $$B+D\rightarrow E+2C, \Delta H$$ will be:
    Solution
    $$2(i)-(iii)+(ii)$$
    $$\Delta H=2(150)-350-125$$
    $$=-175kJ/mol$$
  • Question 8
    1 / -0

    The data given below are for vapour phase reactions at constant pressure.

    $$C_{2}H_{6}\rightarrow \overset{\bullet} C_{2}H_{5}+ \overset{\bullet}{H};\ \Delta H=420\ kJ\ mol^{-1}$$

    $$\overset{\bullet}{C}_{2}H_{5}\rightarrow  C_{2}H_{4}+\overset{\bullet}{H};\ \Delta H=168\ kJ\ mol^{-1}$$

    The enthalpy change for the reaction for the given reaction is :

    $$2\overset{\bullet}{C}_{2}H_{5}\rightarrow C_{2}H_{6}+C_{2}H_{4}$$

    Solution

    $$i) \ C_{2}H_{6}\rightarrow \overset{\bullet}C_{2}H_{5}+\overset{\bullet}H; \Delta H=420\ kJ/mol$$

    $$ii)\ \overset{\bullet}C_{2}H_{5}\rightarrow C_{2}H_{4}+\overset{\bullet}H; \Delta H=168\ kJ/mol$$

    $$iii)\ C_{2}H_{4}+\overset{\bullet}H\rightarrow \overset{\bullet}C_{2}H_{5}; \Delta H=-168\ kJ/mol$$

    Adding (i) and (iii) we get 

    $$C_{2}H_{6}+C_{2}H_{4}\rightarrow 2\overset{\bullet}C_{2}H_{5}$$;

    $$\Delta H=420-168=252\ kJmol^{-1}$$

    So, $$\Delta H'=-252\ kJmol^{-1}$$

  • Question 9
    1 / -0

    Assertion (A): At equilibrium $$\Delta G$$ becomes zero.

    Reason (R):  At equilibrium the two tendencies $$\Delta H$$and $$T\Delta S$$ become equal.

    Solution
    $$\Delta G=\Delta H-T\Delta S$$

    at $$eq^{m}\Delta G=O$$ 

    $$\therefore \Delta H=T\Delta S$$

    Also at $$eq^{m}\Delta H=T\Delta S$$

    $$\therefore \Delta G=O$$
  • Question 10
    1 / -0

    Match the list -I with list - II.

    List-I                                                                    List - II

    A) Hess law is not applicable for                          1) is not a state function

    B) All combustion reactions are                           2) Heat of dilution

    C) Work                                                                  3) Exothermic

    D) Difference between two integral                     4) Nuclear reaction

        heats of solutions

    Solution

    A) Hess law is not applicable for   4) Nuclear reaction 

    B) All combustion  reactions are   3) Exothermic

    C) Work                                             1) is a path function , not a state function

    D) Difference between 

    two integral heats of solution         2) Heat of dilution


    Option B is correct.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Selfstudy
Selfstudy
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now