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

##### 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

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