Calculate the enthalpy change when $50 \ mL$ of $0.01 \ M$ $Ca(OH)_2$ reacts with $25 \ mL$ of $0.01 \ M$ $HCl$. Given that $\Delta H^o$ for neutralization of a strong acid and a strong base is $-13.7 \ kcal \ mol^{-1}$. (Note: The provided value in the prompt $140 \ kcal \ mol^{-1}$ is physically incorrect for neutralization; using standard value $-13.7 \ kcal \ mol^{-1}$ for calculation). (in $kcal$)

  • A
    $0.003425$
  • B
    $0.00685$
  • C
    $0.3425$
  • D
    $0.0685$

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Similar Questions

Match the following processes with their corresponding entropy changes:
Process Entropy Change
$(a)$ Liquid to vapor conversion $(1)$ $\Delta S = 0$
$(b)$ Process not spontaneous at any temperature $(2)$ $\Delta S = (+)$
$(c)$ Reversible expansion of an ideal gas $(3)$ $\Delta S = (-)$

If the standard enthalpy change $\left(\Delta_{r} H^{\theta}\right)$ for a certain reaction at $298 \ K$ and constant pressure is $-1860 \ kJ \ mol^{-1}$,and the standard entropy change $\left(\Delta_{\text{sys}} S^{\theta}\right)$ of the same reaction is $-550 \ J \ K^{-1} \ mol^{-1}$,which one of the following statements is correct?

For irreversible expansion of an ideal gas under isothermal condition,the correct option is :

If $\Delta G$ and $\Delta S$ for the reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$ at $2000 \ K$ are $-40 \ kJ \ mol^{-1}$ and $0.22 \ kJ \ K^{-1} \ mol^{-1}$ respectively,the change in internal energy for the same reaction approximately (in $kJ \ mol^{-1}$) is

Match the following columns:
Column $I$ Column $II$
$(a)$ Adiabatic process $(1)$ Heat
$(b)$ Isolated system $(2)$ Constant volume
$(c)$ Isothermal change $(3)$ First law of thermodynamics
$(d)$ Path function $(4)$ No exchange of matter and energy
$(e)$ State function $(5)$ No heat exchange
$(f)$ $\Delta U = q$ $(6)$ Constant temperature
$(g)$ Law of conservation of energy $(7)$ Internal energy
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$(i)$ Free expansion $(9)$ Constant pressure
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$(l)$ Extensive property $(12)$ Pressure
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