For the reaction 2P + Q rArr R, 8 mol of p and 5 mol of Q will produce:

For the reaction 2P + Q rarr R, 12 mol of P and 8 mol of Q are taken then:

Consider the reaction 2 A+B+3C to P+2Q . Starting with 3 mol of A, 2 mol of B and 6 mol of C, number of moles of the products P and Q would respectively be

The negation of p ^^ (q rarr ~ r) is

for the reaction P+Q rArr R + 2S , initially the concentration of P is equal to that of Q (1 molar) but at equilibrium the concentration of R will be twice of that of P, then the equilibrium constant of the reaction is

In a multistep reaction such as A + B rarr Q rarr C . The potential energy diagram is shown below. What is E_(a) for the reaction Q rarr C ? (a) 3 kcal mol^(-1) (b) 5 kcal mol^(-1) (c) 8 kcal mol^(-1) (d) 11 kcal mol^(-1)

The negation of p ^^(q rarr ~r) is

For the reaction 2P + Q-------> P_(2)Q , the rate of formation of P_(2)Q is 0.24 mol dm^(-3)s^(-1) . Then the rates of disappearance of P and Q respectively are

For a reaction, P+Q rarr R+S . The value of DeltaH^(@) is -"30 kJ mol"^(-1) and DeltaS" is "-"100 J K"^(-1)"mol"^(-1) . At what temperature the reaction will be at equilibrium?

For a reaction, P+Q rarr R+S . The value of DeltaH^(@) is -"30 kJ mol"^(-1) and DeltaS" is "-"100 J K"^(-1)"mol"^(-1) . At what temperature the reaction will be at equilibrium?