Catalysts lower activation energies equally for forward and backward reactions without altering ∆H
Step 1: Activation energy for the uncatalyzed backward reaction:
\( E_a(\text{backward}) = E_a(\text{forward}) - \Delta H \)
\( E_a(\text{backward}) = 300 - 20 = 280 \, \text{kJ/mol} \)
Step 2: Using the given temperatures and equal rates, calculate \( E_a(\text{forward, catalyzed}) \):
\( \frac{E_a(\text{forward, catalyzed})}{E_a(\text{forward, uncatalyzed})} = \frac{T_c}{T_u} \)
\( \frac{E_a(\text{forward, catalyzed})}{300} = \frac{300}{600} \)
\( E_a(\text{forward, catalyzed}) = 150 \, \text{kJ/mol} \)
Step 3: Calculate \( E_a(\text{backward, catalyzed}) \):
\( E_a(\text{backward, catalyzed}) = E_a(\text{forward, catalyzed}) - \Delta H \)
\( E_a(\text{backward, catalyzed}) = 150 - 20 = 130 \, \text{kJ/mol} \)
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to:
The output of the circuit is low (zero) for:

(A) \( X = 0, Y = 0 \)
(B) \( X = 0, Y = 1 \)
(C) \( X = 1, Y = 0 \)
(D) \( X = 1, Y = 1 \)
Choose the correct answer from the options given below:
The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are
A. $ Cr^{2+} $
B. $ Fe^{2+} $
C. $ Fe^{3+} $
D. $ Co^{2+} $
E. $ Mn^{2+} $
Choose the correct answer from the options given below
Which of the following circuits has the same output as that of the given circuit?
