If you wanted to carry out self-propagating high-temperature synthesis, you could in many cases mix the synthesis feedstocks you desire to heat up with a redox fuel mix that will produce the desired heat, with a grain size chosen to produce the desired reaction rate.
The conventional reducer for this sort of thing is aluminum, with its oxide’s enthalpy of formation of -1675.7 kJ/mol (-559/mol O). Magnesium at -601.6 (-601.6), beryllium at -599 (-599), and lithium at -595.8 (-595.8) excel it. Other candidate reducers include zirconium at -1080 (-540), silicon at -911 (-456), zinc at -350.5 (-350.5), and boron at -1254 (-418). All of these have extremely stable and unreactive oxides, at least at room temperature.
The conventional oxidizer is hematite at -824.2 kJ/mol (-274.7), yielding 283.8 kJ per mole of O1 when oxidizing aluminum. This is 5.33 kJ per gram of hematite at 159.687 g/mol. At 5.25 g/cc, that’s 27.99 kJ/cc of hematite. But of course you also need 2 mol Al (26.9815 g/mol, so 53.96 g) at 2.70 g/cc (20.0 cc) to react with each mol (30.4 cc) of Fe2O3, so you actually only get 3.98 kJ/g of mix or 16.9 kJ/cc.
Hematite is abundant and cheap, but it has several disadvantages. Though in pure form it gets hot enough to boil the iron, you can’t dilute this fuel mix very far with your actual desired reagents before the reaction stops being self-propagating, maybe a factor of 2 or 3. And the metallic iron produced is itself fairly reactive. Consider, by contrast, black cupric oxide (tenorite) at -156 (-156), thus yielding 403 kJ/mol; at 79.545 g/mol, you get 5.1 kJ/g of oxidizer, and the copper produced is considerably less reactive than iron (though easier to boil if the fuel is dangerously concentrated). Other promising alternative oxidizers include NiO at -240 (-240), chromium trioxide at -589.3 (-196.4), lead dioxide at -274.47 (-137.24), manganese dioxide at -520 (-260), and potassium permanganate at -813.4 (-203.4).
Consider a stoichiometric mix of lead dioxide with magnesium. We need 2 mol of magnesium (24.305 g/mol, so 48.610 g, occupying 27.895 cc at 1.737 g/cc) per mol (239.1988 g) of lead dioxide, occupying at 25.5 cc at 9.38 g/cc, for a total of 287.808 g in 53.4 cc. Burning it produces 2 * 601.6 - 274.47 = 928.73 kJ (464.37 kJ per mol of O1), which is 17.4 kJ/cc or 3.23 kJ/g.
This is sort of disappointing, although it is at least a little more heat per volume. I feel like I must have miscalculated something; the enthalpy yield per oxygen atom is double the conventional system, but the yield per cc is only slightly higher, and per g it’s actually lower. I guess the lead is just too heavy and bulky?
How about lithium with manganese dioxide? You need 4 mol lithium (6.94 g/mol, so 27.8 g, occupying 52.0 cc at 0.534 g/cc) per mol of MnO2 (86.9368 g, occupying 17.30 cc at 5.026 g/cc), for a total of 114.7 g in 69.3 cc. It burns to metallic manganese and 2 mol of Li2O and 2 * 595.8 kJ - 520 kJ = 672 kJ, 9.69 kJ/cc or 5.86 kJ/g. Super disappointing!
Still, not so disappointing as to be useless for boosting SHS. Even fairly small admixtures of these boosters may be adequate to permit SHS of feedstocks that would be hopeless on their own.