1. Tardigrade
  2. Question
  3. Chemistry
  4. Based on the following thermochemical equations H2O(g) + C (s) → CO(g) + H2 (g); Δ H= 131 kJ CO(g) + 1/2 O2(g)→ CO2(g); Δ H= -2 82 kJ H2(g) + 1/2 O2(g)→ H2O(g); Δ H= - 242 kJ C(s) + O2(g)→ CO2(g); Δ H=X kJ The value of X will be

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Q. Based on the following thermochemical equations
$H_2O\left(g\right) + C \left(s\right) \to CO\left(g\right) + H_2 \left(g\right); \Delta H= 131 kJ$
$CO\left(g\right) + 1/2 O_2\left(g\right)\to CO_2\left(g\right); \Delta H= -2 82 kJ$
$H_2\left(g\right) + 1/2 O_2\left(g\right)\to H_2O\left(g\right); \Delta H= - 242 kJ$
$C\left(s\right) + O_2\left(g\right)\to CO_2\left(g\right); \Delta H=X kJ$
The value of $X$ will be

Thermodynamics

Solution:

The reaction $C\left(s\right) + O_2\left(g\right)\to CO_2\left(g\right)$
is the result of the addition of the given three reactions
$\therefore \Delta H = X= 131 + \left(-282\right) + \left(-242\right)$
$= -393 kJ$