Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
- A$-393$
- B$+393$
- C$+655$
- D$-655$
Explore More
Similar Questions
Consider the following data :
Heat of formation of $CO_{2(g)} = -393.5 \ kJ \ mol^{-1}$
Heat of formation of $H_2O_{(l)} = -286.0 \ kJ \ mol^{-1}$
Heat of combustion of benzene $= -3267.0 \ kJ \ mol^{-1}$
The heat of formation of benzene is $........... \ kJ \ mol^{-1}$. $(Nearest \ integer)$
Heat of formation of $CO_{2(g)} = -393.5 \ kJ \ mol^{-1}$
Heat of formation of $H_2O_{(l)} = -286.0 \ kJ \ mol^{-1}$
Heat of combustion of benzene $= -3267.0 \ kJ \ mol^{-1}$
The heat of formation of benzene is $........... \ kJ \ mol^{-1}$. $(Nearest \ integer)$
For the reaction,$2 H_{2(g)} + O_{2(g)} \longrightarrow 2 H_{2}O_{(g)}$,$\Delta H^{\circ} = -573.2 \ kJ$. The heat of decomposition of water per mole is
Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)
DifficultTS EAMCET 2007
View SolutionCalculate 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 $kJ$)
The reaction for the formation of $NaCl$ is:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo