From the thermo chemical reactions.
`C_(graphite)` + `1/2``O_2` `to` `CO` `Delta` H = -110.5 KJ
`CO` + `1/2``O_2` `to` `CO_2`; `Delta`H = -283.2 KJ
`Delta` H For the reaction, `C_(graphite)` + `O_2` `to` `CO_2` is

Form the thermochemical reactions. C(graphite) +1//2 O_(2)(g) to CO(g), triangleH=-110.5kJ CO(g)+1//2O_(2)(g) to CO_(2)(g), triangleH=-283.2kJ The enthalpy ofreaction of C(graphite) +O_(2)(g) to CO_(2)" is "

Complete the chemical equations : CO_2+ …… to C_6H_12O_6 + …. + H_2O

C(s)+(1)/(2)O_(2)(g)to CO(g), Delta H =- 42 kJ/mole CO(g)+(1)/(2)O_(2)(g)to CO_(2)(g) , Delta H =-24 kJ/mole The heat of formation of CO_(2) is

Balance the equations : C_2H_4 + O_2 to CO_2 + H_2O

For the following reaction in gaseous phase, CO + (1)/(2) O_2 to CO_2 ,(K_c)/( K_p) is

According to the following reaction C(S)+1//2O_(2)(g)rarr CO(g), Delta H=-26.4 Kcal

Balance the equations : C_2H_4+O_2 to CO_2 + H_2O

Calculate enthalpy of formation of methane (CH_4) from the following data : (i) C(s) + O_(2)(g) to CO_(2) (g) , Delta_rH^(@) = -393.5 KJ mol^(-1) (ii) H_2(g) + 1/2 O_(2)(g) to H_(2)O(l) , Deta_r H^(@) = -285.5 kJ mol^(-1) (iii) CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l), Delta_(r)H^(@) = -890.3 kJ mol^(-1) .

Consider the following reactions: I. C(s) + 1/2 O_(2)(g) rarr CO(g)," "Delta H_(1) = x_(1) II. CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " "Delta H_(2) = x_(2) III. C(s) + CO_(2)(g) rarr 2CO(g)," "Delta H_(3) = x_(3) Select the correct statements:

Based on the following reaction C(graphite) + O_2(g)rarrCO_2(g),DeltaH=-394KJ//mol...(i) 2CO(g)+O_2(g)rarr2CO_2(g),DeltaH=-569KJ//mol...(ii) The heat of formation of CO will be