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

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

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

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

In the reaction, P + Q rarr R + S, the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in figure. The overall order of the reaction 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 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)

If the inverse of implication p rarr q is defined as ~p rarr ~q , then the inverse of the proposition (p ^^ ~q) rarr r is not equivalent to : I : ~r rarr p vv q II : ~p vv ~q rarr ~r III : r rarr p ^^ ~q The true statements in the above are :