#### Answer

b)
$[ A ] = 0.41 M$
$[ B ] = 0.41 M$
$[ C ] = 0.054 M$

#### Work Step by Step

- The new reaction is:
$$2C \leftrightharpoons A + B$$
So, the new equilibrium constant is:
$$K_c = \frac{1}{1.7 \times 10^{-2}} = 59$$
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ A ][ B ]}{[ C ]^{ 2 }}$$
2. At equilibrium, these are the concentrations of each compound:
$[ C ] = 0.875 \space M - 2x$
$[ A ] = x$
$[ B ] = x$
$$59 = \frac{(x)(x)}{(0.875-2x)^2}$$
x $\approx$ 0.4107 (in order to get a positive concentration of C)
3.
$[ C ] = 0.875 \space M - 2(0.4107) M= 0.054 M$
$[ A ] = 0.41 M$
$[ B ] = 0.41 M$