`2NO_2(g) harr N_2O_4(g): DeltaH =-52.7 kJmol_1` What happens when `N_2O_4` is removed from the reaction medium?

2NO_2(g) harr N_2O_4(g): DeltaH =-52.7 kJmol_1 What change will happen if we increase the temperature?

2NO_2(g) harr N_2O_4(g): DeltaH =-52.7 kJmol_1 What is the effect of increase in pressure in the above equilibrium?

Consider the reaction 4NO_(2(g)) + O _(2 (g)) to 2 N _(2 ) O _(5(g)) , Delta H =-111 kJ If N _(2 ) O _(5 (s)) is formed in the above reaction Delta H will be (Given Delta H for sublimation of N_(2) O_(5) is 54 kJ)

Ethalpies of formation of CO(g), CO_2(g) ,N_2O(g) and N_2O_4 (g) are -110,-393,81 and 9.7kJ mol^-1 respectively. Find the value of Delta_rH from the reaction N_2O_4(g)+3CO(g)rarrN_2O(g)+3 CO_2(g)

a) Find the enthalpy of the reaction, C(graphite) + O_2(g)rightarrowCO_2(g) Given, i) C(graphite) +frac{1}{2}O_2(g)rightarrowCO(g),DeltaH=-110.5kJmol^-1 ii) CO(g) +frac{1}{2}O_2(g)rightarrowCO_2(g),DeltaH=-283.0kJmol^-1 b) Melting of ice is a spontaneousnprocess. What are the criteria for spontaneity of a process?

Observe the equation of thechemical reactions given below. NaOH (aq) + HCI (aq) harr 2NaCI (aq)+H_2O(I) , N_2(g)+O_2(g)harr2NO(g) , Zn(s)+2HCIrarrZnCI_2(aq)+H_2(g) , 2SO_2(g)+O_2(g)harr2SO_3(g) : In this reaction what changes will be made in pressure for increasing forward reaction? Why?