Which is the best definition of “heat of neutralization”?
- AThe heat set free when one gram molecule of a base is neutralized by one gram molecule of an acid in dilute solution at a stated temperature.
- BThe heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperature.
- CThe heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperature.
- DThe heat set free when one gram equivalent of an acid is neutralized by one gram equivalent of a base in dilute solution at a stated temperature.
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Similar Questions
The enthalpies of combustion of carbon and carbon monoxide are $-393.5 \, kJ$ and $-283 \, kJ$ respectively. Calculate the enthalpy of formation of carbon monoxide in $kJ$. (in $.5$)
Medium
View SolutionThe following thermochemical reactions are given:
$M + \frac{1}{2} O_2 \to MO + 351.4 \ kJ$ $(i)$
$X + \frac{1}{2} O_2 \to XO + 90.8 \ kJ$ $(ii)$
It follows that the heat of reaction for the following process $M + XO \rightleftharpoons MO + X$ is given by $... \ kJ$.
$M + \frac{1}{2} O_2 \to MO + 351.4 \ kJ$ $(i)$
$X + \frac{1}{2} O_2 \to XO + 90.8 \ kJ$ $(ii)$
It follows that the heat of reaction for the following process $M + XO \rightleftharpoons MO + X$ is given by $... \ kJ$.
Medium
View SolutionFrom the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:
Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)
According to the following figure,the magnitude of the enthalpy change of the reaction $A+B \rightarrow M+N$ in $kJ \ mol^{-1}$ is equal to ...... .
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