Nitroglycerine $(MW = 227.1)$ detonates according to the following equation:
$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$
The standard molar enthalpies of formation,$\Delta H_{f}^{\circ}$ for all the compounds are given below:
$\Delta H_{f}^{\circ}[C_{3}H_{5}(NO_{3})_{3}] = -364 \, kJ/mol$
$\Delta H_{f}^{\circ}[CO_{2(g)}] = -393.5 \, kJ/mol$
$\Delta H_{f}^{\circ}[H_{2}O_{(g)}] = -241.8 \, kJ/mol$
$\Delta H_{f}^{\circ}[N_{2(g)}] = 0 \, kJ/mol$
$\Delta H_{f}^{\circ}[O_{2(g)}] = 0 \, kJ/mol$
The enthalpy change when $10 \, g$ of nitroglycerine is detonated is $..... \, kJ$.
$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$
The standard molar enthalpies of formation,$\Delta H_{f}^{\circ}$ for all the compounds are given below:
$\Delta H_{f}^{\circ}[C_{3}H_{5}(NO_{3})_{3}] = -364 \, kJ/mol$
$\Delta H_{f}^{\circ}[CO_{2(g)}] = -393.5 \, kJ/mol$
$\Delta H_{f}^{\circ}[H_{2}O_{(g)}] = -241.8 \, kJ/mol$
$\Delta H_{f}^{\circ}[N_{2(g)}] = 0 \, kJ/mol$
$\Delta H_{f}^{\circ}[O_{2(g)}] = 0 \, kJ/mol$
The enthalpy change when $10 \, g$ of nitroglycerine is detonated is $..... \, kJ$.
- A$-100.5$
- B$-62.5$
- C$-80.3$
- D$-74.9$
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The heat of neutralisation will be highest in
Easy
View SolutionFor the following reaction,
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)
Medium
View SolutionCalculate the heat of combustion (in $kJ$) of methane from the following data:
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
DifficultTS EAMCET 2002
View SolutionFill in the blanks:
$(i)$ The enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states is called the ....... of that compound.
$(ii)$ The total heat change in a chemical reaction is equal to the algebraic sum of the heat changes of the individual steps of the reaction. This law was given by .......... .
$(iii)$ $A$ process during which there is no exchange of heat between the system and the surroundings is called an ........... process.
$(i)$ The enthalpy change that occurs when one mole of a compound is formed from its constituent elements in their standard states is called the ....... of that compound.
$(ii)$ The total heat change in a chemical reaction is equal to the algebraic sum of the heat changes of the individual steps of the reaction. This law was given by .......... .
$(iii)$ $A$ process during which there is no exchange of heat between the system and the surroundings is called an ........... process.
Medium
View SolutionUse the data from the table to estimate the enthalpy of formation of $CH_3CHO$.
Bond Bond Enthalpy $(kJ \ mol^{-1})$ Enthalpy of formation $(kJ \ mol^{-1})$ $C-H$ $400$ $C(g): 700$ $C-C$ $350$ $H(g): 200$ $C=O$ $700$ $O(g): 250$
| Bond | Bond Enthalpy $(kJ \ mol^{-1})$ | Enthalpy of formation $(kJ \ mol^{-1})$ |
|---|---|---|
| $C-H$ | $400$ | $C(g): 700$ |
| $C-C$ | $350$ | $H(g): 200$ |
| $C=O$ | $700$ | $O(g): 250$ |
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