The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be

(i) \( -74.8 kJmol^{-1}\)

(ii) \( -52.27 kJmol^{-1}\)

(iii) \( +74.8 kJmol^{-1}\)

(iv) \( +52.26 kJmol^{-1}\)

Asked by Abhisek | 1 year ago |  77

1 Answer

Solution :-

(i) \( -74.8kJmol^{-1}\)

1. CH4(g)  + 2O2(g) → CO2(g) + 2H2O(g)

\( \Delta H = -890.3kJmol^{-1}\)

 

(ii) C(s) + O2(g) → CO2(g)

\( \Delta H = -393.5kJmol^{-1} \)

 

(iii) 2H2(g) + O2(g) → 2H2O(g)

\( \Delta H = -285.8kJmol^{-1}\)

C(s) + 2H2(g) → CH4(g)

\( \Delta _{f}H_{CH_{4}} = \Delta _{c}H_{c}+ 2\Delta _{f}H_{H_{2}}​​ – \Delta _{f}H_{CO_{2}} \)

= [ -393.5 +2(-285.8) – (-890.3)]\( kJmol^{-1}\)

= -74.8\( kJmol^{-1}\)

Answered by Pragya Singh | 1 year ago

Related Questions

Calculate the entropy change in surroundings when 1.00 mol of \( H_2O\)(l) is formed under standard conditions. \( Δ_f H^ Θ= –286 kJ mol^{–1}.\)

Class 11 Chemistry Thermodynamics View Answer

Comment on the thermodynamic stability of NO(g), given,

(\( \dfrac{1}{2}\))N2(g) + (\( \dfrac{1}{2}\))O2(g) → NO(g)\( \Delta _{r}H^{\Theta } = 90kJmol^{-1}\)

NO(g) + (\( \dfrac{1}{2}\))O2(g) → NO2(g)\( \Delta _{r}H^{\Theta } = -74kJmol^{-1}\)

Class 11 Chemistry Thermodynamics View Answer

The equilibrium constant for a reaction is 10. What will be the value of ∆G0? R = 8.314 JK–1 mol–1, T = 300 K.

Class 11 Chemistry Thermodynamics View Answer

For the reaction

2A(g) + B(g)→ 2D(g)

\( \Delta U^{\Theta } = -10.5 kJ \;and \;\Delta S^{\Theta } = -44.1JK^{-1}\)

Calculate \( \Delta G^{\Theta }\) for the reaction, and predict whether the reaction may occur spontaneously.

Class 11 Chemistry Thermodynamics View Answer

For the reaction

2Cl(g)→ Cl2(g)

What are the signs of \( \Delta S\; and \;\Delta H\)?

Class 11 Chemistry Thermodynamics View Answer