`{:(,"Column-I",,"Column-II",),(,,,("mass of product"),),((A),2H_(2)+O_(2)rarr2H_(2)O,(p),1.028 g,),(,lg" "lg,,,),((B),3H_(2) +N_(2) rarr2NH_(3),(q),1.333 g,),(,lg " " lg,,,),((C),H_(2) + CI_(2) rarr 2HCI ,(r),1.125 g,),(,lg " "lg ,,,),((D),2H_(2) + C rarrCH_(4),(s),1.214 g,),(,lg " "lg,,,):}`

H_(2)(g)+(1)/(2)O_(2)(g)rarr2H_(2)O(l), DeltaH =- 286 kJ 2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l)……………kJ(+-?)

Match the following {:(,"Column -I",,"Column -II"),((A),underset(3g)(2H_(2))+underset(22.66g)(O_(2))rarr 2H_(2)O,(p),25.5 "g product"),((B),underset(21 g)(N_(2))+underset(5 g)(3H_(2))rarr 2 NH_(3),(q),H_(2) "is present in excess"),((C ),underset(1g)(H_(2))+underset(40 g)(Cl_(2))rarr 2 HCl,(r ),H_(2) "is the limiting reagent"),((D),underset(20g)(C )+underset(6.375g)(2H_(2))rarr CH_(4),(s),36.5"g product"):}

Identify the type of reaction 2H_(2(g))+O_(2(g))rarr2H_2O_((I))

{:(,"Column-I(Reactions)",,,"Column-II(Product)"),((A),"Borax",overset(Delta)rarr,(p),"BN"),((B),B_(2)H_(6)+H_(2)O,overset(Delta)rarr,(q),B_(2)H_(6)),((C),B_(2)H_(6)NH_(3)("Excess"),overset(Delta)rarr,(r),H_(3)BO_(3)),((D),BCl_(3)+LiAlH_(4),overset(Delta)rarr,(s),NaBO_(2)+B_(2)O_(3)):}

Calculate in kJ for the following reaction : C(g) + O_(2)(g) rarr CO_(2)(g) Given that, H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ

Calculate in kJ for the following reaction : C(g) + O_(2)(g) rarr CO_(2)(g) Given that, H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ

Diborane is a potential rocket fuel which undergoes combustion according to the reaction, B_(2)H_(6) (g) + 3O_(2) (g) rarr B_(2) O_(3) (s) + 3H_(2) O (g) From the following data, calculate the enthalpy change for the combustion of diborane: (i) 2 B (s) + ((3)/(2)) O_(2) (g) rarr B_(2) O_(3) (s) , Delta H = 0 1273 k J//mol (ii) H_(2) (g) + ((1)/(2)) O_(2) (g) rarr H_(2) O (l), Delta H = -286 kJ//mol (iii) H_(2) O (l) rarr H_(2)O (g) Delta H = 44 kJ//mol (iv) 2B (s) + 3H_(2) (G) rarr B_(2) H_(6) (g), Delta H = 36 kJ//mol

In the following process/es Delta H lt Delta U (A) C(s) + O_(2)(g) rarr CO_(2)(g) (B) CO(g)+(1)/(2)O_(2)(g) rarr CO_(2)(g) (C) N_(2)(g)+3H_(2)(g) rarr 2NH_(3)(g) (D) CaCO_(3)(s) rarr CaO(s) + CO_(2)(g)

Calculate Delta H^(theta) for the reaction : H_(2)O (l) rarr H_(2)(g) + (1)/(2) O_(2) (g) Given 2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) Delta H = -571.6 kJ